Pre-Algebra Examples

$\begin{array}{c} x & y \\ - 2 & \frac{1}{81} \\ - 1 & \frac{1}{9} \\ 0 & 1 \\ 1 & 9 \\ 2 & 81 \end{array}$

Step 1

Check if the function rule is linear.

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Step 1.1

To find if the table follows a function rule, check to see if the values follow the linear form $y = a x + b$ .

$y = a x + b$

Step 1.2

Build a set of equations from the table such that $y = a x + b$ .

$\frac{1}{81} = a (- 2) + b \frac{1}{9} = a (- 1) + b 1 = a (0) + b 9 = a (1) + b 81 = a (2) + b$

Step 1.3

Calculate the values of $a$ and $b$ .

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Step 1.3.1

Rewrite the equation as $b = 1$ .

$b = 1$

$\frac{1}{81} = a (- 2) + b$

$\frac{1}{9} = a (- 1) + b$

$9 = a + b$

$81 = a (2) + b$

Step 1.3.2

Replace all occurrences of $b$ with $1$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $b$ in $\frac{1}{81} = a (- 2) + b$ with $1$ .

$\frac{1}{81} = a (- 2) + 1$

$b = 1$

$\frac{1}{9} = a (- 1) + b$

$9 = a + b$

$81 = a (2) + b$

Step 1.3.2.2

Simplify $\frac{1}{81} = a (- 2) + 1$ .

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Step 1.3.2.2.1

Simplify the left side.

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Step 1.3.2.2.1.1

Remove parentheses.

$\frac{1}{81} = a (- 2) + 1$

$b = 1$

$\frac{1}{9} = a (- 1) + b$

$9 = a + b$

$81 = a (2) + b$

$\frac{1}{81} = a (- 2) + 1$

$b = 1$

$\frac{1}{9} = a (- 1) + b$

$9 = a + b$

$81 = a (2) + b$

Step 1.3.2.2.2

Simplify the right side.

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Step 1.3.2.2.2.1

Move $- 2$ to the left of $a$ .

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$\frac{1}{9} = a (- 1) + b$

$9 = a + b$

$81 = a (2) + b$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$\frac{1}{9} = a (- 1) + b$

$9 = a + b$

$81 = a (2) + b$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$\frac{1}{9} = a (- 1) + b$

$9 = a + b$

$81 = a (2) + b$

Step 1.3.2.3

Replace all occurrences of $b$ in $\frac{1}{9} = a (- 1) + b$ with $1$ .

$\frac{1}{9} = a (- 1) + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$9 = a + b$

$81 = a (2) + b$

Step 1.3.2.4

Simplify $\frac{1}{9} = a (- 1) + 1$ .

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Step 1.3.2.4.1

Simplify the left side.

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Step 1.3.2.4.1.1

Remove parentheses.

$\frac{1}{9} = a (- 1) + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$9 = a + b$

$81 = a (2) + b$

$\frac{1}{9} = a (- 1) + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$9 = a + b$

$81 = a (2) + b$

Step 1.3.2.4.2

Simplify the right side.

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Step 1.3.2.4.2.1

Simplify each term.

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Step 1.3.2.4.2.1.1

Move $- 1$ to the left of $a$ .

$\frac{1}{9} = - 1 a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$9 = a + b$

$81 = a (2) + b$

Step 1.3.2.4.2.1.2

Rewrite $- 1 a$ as $- a$ .

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$9 = a + b$

$81 = a (2) + b$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$9 = a + b$

$81 = a (2) + b$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$9 = a + b$

$81 = a (2) + b$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$9 = a + b$

$81 = a (2) + b$

Step 1.3.2.5

Replace all occurrences of $b$ in $9 = a + b$ with $1$ .

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$81 = a (2) + b$

Step 1.3.2.6

Simplify the left side.

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Step 1.3.2.6.1

Remove parentheses.

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$81 = a (2) + b$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$81 = a (2) + b$

Step 1.3.2.7

Replace all occurrences of $b$ in $81 = a (2) + b$ with $1$ .

$81 = a (2) + 1$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.2.8

Simplify $81 = a (2) + 1$ .

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Step 1.3.2.8.1

Simplify the left side.

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Step 1.3.2.8.1.1

Remove parentheses.

$81 = a (2) + 1$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$81 = a (2) + 1$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.2.8.2

Simplify the right side.

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Step 1.3.2.8.2.1

Move $2$ to the left of $a$ .

$81 = 2 a + 1$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$81 = 2 a + 1$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$81 = 2 a + 1$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$81 = 2 a + 1$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.3

Solve for $a$ in $81 = 2 a + 1$ .

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Step 1.3.3.1

Rewrite the equation as $2 a + 1 = 81$ .

$2 a + 1 = 81$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.3.2

Move all terms not containing $a$ to the right side of the equation.

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Step 1.3.3.2.1

Subtract $1$ from both sides of the equation.

$2 a = 81 - 1$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.3.2.2

Subtract $1$ from $81$ .

$2 a = 80$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$2 a = 80$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.3.3

Divide each term in $2 a = 80$ by $2$ and simplify.

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Step 1.3.3.3.1

Divide each term in $2 a = 80$ by $2$ .

$\frac{2 a}{2} = \frac{80}{2}$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.3.3.2

Simplify the left side.

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Step 1.3.3.3.2.1

Cancel the common factor of $2$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{2 a}{2} = \frac{80}{2}$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{80}{2}$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$a = \frac{80}{2}$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$a = \frac{80}{2}$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide $80$ by $2$ .

$a = 40$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$a = 40$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$a = 40$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$a = 40$

$9 = a + 1$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.4

Replace all occurrences of $a$ with $40$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $a$ in $9 = a + 1$ with $40$ .

$9 = (40) + 1$

$a = 40$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Add $40$ and $1$ .

$9 = 41$

$a = 40$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$9 = 41$

$a = 40$

$\frac{1}{9} = - a + 1$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.4.3

Replace all occurrences of $a$ in $\frac{1}{9} = - a + 1$ with $40$ .

$\frac{1}{9} = - (40) + 1$

$9 = 41$

$a = 40$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.4.4

Simplify the right side.

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Step 1.3.4.4.1

Simplify $- (40) + 1$ .

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Step 1.3.4.4.1.1

Multiply $- 1$ by $40$ .

$\frac{1}{9} = - 40 + 1$

$9 = 41$

$a = 40$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.4.4.1.2

Add $- 40$ and $1$ .

$\frac{1}{9} = - 39$

$9 = 41$

$a = 40$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$\frac{1}{9} = - 39$

$9 = 41$

$a = 40$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

$\frac{1}{9} = - 39$

$9 = 41$

$a = 40$

$\frac{1}{81} = - 2 a + 1$

$b = 1$

Step 1.3.4.5

Replace all occurrences of $a$ in $\frac{1}{81} = - 2 a + 1$ with $40$ .

$\frac{1}{81} = - 2 \cdot 40 + 1$

$\frac{1}{9} = - 39$

$9 = 41$

$a = 40$

$b = 1$

Step 1.3.4.6

Simplify the right side.

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Step 1.3.4.6.1

Simplify $- 2 \cdot 40 + 1$ .

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Step 1.3.4.6.1.1

Multiply $- 2$ by $40$ .

$\frac{1}{81} = - 80 + 1$

$\frac{1}{9} = - 39$

$9 = 41$

$a = 40$

$b = 1$

Step 1.3.4.6.1.2

Add $- 80$ and $1$ .

$\frac{1}{81} = - 79$

$\frac{1}{9} = - 39$

$9 = 41$

$a = 40$

$b = 1$

$\frac{1}{81} = - 79$

$\frac{1}{9} = - 39$

$9 = 41$

$a = 40$

$b = 1$

$\frac{1}{81} = - 79$

$\frac{1}{9} = - 39$

$9 = 41$

$a = 40$

$b = 1$

$\frac{1}{81} = - 79$

$\frac{1}{9} = - 39$

$9 = 41$

$a = 40$

$b = 1$

Step 1.3.5

Since $\frac{1}{81} = - 79$ is not true, there is no solution.

No solution

Step 1.4

Since $y \neq y$ for the corresponding $x$ values, the function is not linear.

The function is not linear

Step 2

Check if the function rule is quadratic.

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Step 2.1

To find if the table follows a function rule, check whether the function rule could follow the form $y = a x^{2} + b x + c$ .

$y = a x^{2} + b x + c$

Step 2.2

Build a set of $3$ equations from the table such that $y = a x^{2} + b x + c$ .

Step 2.3

Calculate the values of $a$ , $b$ , and $c$ .

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Step 2.3.1

Solve for $c$ in $1 = a \cdot 0^{2} + c$ .

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Step 2.3.1.1

Rewrite the equation as $a \cdot 0^{2} + c = 1$ .

$a \cdot 0^{2} + c = 1$

$\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + c$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.1.2

Simplify $a \cdot 0^{2} + c$ .

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Step 2.3.1.2.1

Simplify each term.

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Step 2.3.1.2.1.1

Raising $0$ to any positive power yields $0$ .

$a \cdot 0 + c = 1$

$\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + c$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.1.2.1.2

Multiply $a$ by $0$ .

$0 + c = 1$

$\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + c$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

$0 + c = 1$

$\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + c$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.1.2.2

Add $0$ and $c$ .

$c = 1$

$\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + c$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

$c = 1$

$\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + c$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

$c = 1$

$\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + c$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2

Replace all occurrences of $c$ with $1$ in each equation.

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Step 2.3.2.1

Replace all occurrences of $c$ in $\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + c$ with $1$ .

$\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + 1$

$c = 1$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.2

Simplify $\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + 1$ .

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Step 2.3.2.2.1

Simplify the left side.

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Step 2.3.2.2.1.1

Remove parentheses.

$\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + 1$

$c = 1$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

$\frac{1}{81} = a {(- 2)}^{2} + b (- 2) + 1$

$c = 1$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.2.2

Simplify the right side.

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Step 2.3.2.2.2.1

Simplify each term.

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Step 2.3.2.2.2.1.1

Raise $- 2$ to the power of $2$ .

$\frac{1}{81} = a \cdot 4 + b (- 2) + 1$

$c = 1$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.2.2.1.2

Move $4$ to the left of $a$ .

$\frac{1}{81} = 4 \cdot a + b (- 2) + 1$

$c = 1$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.2.2.1.3

Move $- 2$ to the left of $b$ .

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.3

Replace all occurrences of $c$ in $\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + c$ with $1$ .

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.4

Simplify $\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + 1$ .

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Step 2.3.2.4.1

Simplify the left side.

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Step 2.3.2.4.1.1

Remove parentheses.

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

$\frac{1}{9} = a {(- 1)}^{2} + b (- 1) + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.4.2

Simplify the right side.

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Step 2.3.2.4.2.1

Simplify each term.

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Step 2.3.2.4.2.1.1

Raise $- 1$ to the power of $2$ .

$\frac{1}{9} = a \cdot 1 + b (- 1) + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.4.2.1.2

Multiply $a$ by $1$ .

$\frac{1}{9} = a + b (- 1) + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.4.2.1.3

Move $- 1$ to the left of $b$ .

$\frac{1}{9} = a - 1 b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.4.2.1.4

Rewrite $- 1 b$ as $- b$ .

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$9 = a + b + c$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.5

Replace all occurrences of $c$ in $9 = a + b + c$ with $1$ .

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.6

Simplify the left side.

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Step 2.3.2.6.1

Remove parentheses.

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = a \cdot 2^{2} + b (2) + c$

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = a \cdot 2^{2} + b (2) + c$

Step 2.3.2.7

Replace all occurrences of $c$ in $81 = a \cdot 2^{2} + b (2) + c$ with $1$ .

$81 = a \cdot 2^{2} + b (2) + 1$

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.2.8

Simplify $81 = a \cdot 2^{2} + b (2) + 1$ .

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Step 2.3.2.8.1

Simplify the left side.

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Step 2.3.2.8.1.1

Remove parentheses.

$81 = a \cdot 2^{2} + b (2) + 1$

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = a \cdot 2^{2} + b (2) + 1$

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.2.8.2

Simplify the right side.

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Step 2.3.2.8.2.1

Simplify each term.

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Step 2.3.2.8.2.1.1

Raise $2$ to the power of $2$ .

$81 = a \cdot 4 + b (2) + 1$

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.2.8.2.1.2

Move $4$ to the left of $a$ .

$81 = 4 \cdot a + b (2) + 1$

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.2.8.2.1.3

Move $2$ to the left of $b$ .

$81 = 4 a + 2 b + 1$

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = 4 a + 2 b + 1$

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = 4 a + 2 b + 1$

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = 4 a + 2 b + 1$

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = 4 a + 2 b + 1$

$9 = a + b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.3

Solve for $a$ in $9 = a + b + 1$ .

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Step 2.3.3.1

Rewrite the equation as $a + b + 1 = 9$ .

$a + b + 1 = 9$

$81 = 4 a + 2 b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.3.2

Move all terms not containing $a$ to the right side of the equation.

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Step 2.3.3.2.1

Subtract $b$ from both sides of the equation.

$a + 1 = 9 - b$

$81 = 4 a + 2 b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.3.2.2

Subtract $1$ from both sides of the equation.

$a = 9 - b - 1$

$81 = 4 a + 2 b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.3.2.3

Subtract $1$ from $9$ .

$a = - b + 8$

$81 = 4 a + 2 b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$a = - b + 8$

$81 = 4 a + 2 b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$a = - b + 8$

$81 = 4 a + 2 b + 1$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.4

Replace all occurrences of $a$ with $- b + 8$ in each equation.

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Step 2.3.4.1

Replace all occurrences of $a$ in $81 = 4 a + 2 b + 1$ with $- b + 8$ .

$81 = 4 (- b + 8) + 2 b + 1$

$a = - b + 8$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.4.2

Simplify the right side.

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Step 2.3.4.2.1

Simplify $4 (- b + 8) + 2 b + 1$ .

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Step 2.3.4.2.1.1

Simplify each term.

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Step 2.3.4.2.1.1.1

Apply the distributive property.

$81 = 4 (- b) + 4 \cdot 8 + 2 b + 1$

$a = - b + 8$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.4.2.1.1.2

Multiply $- 1$ by $4$ .

$81 = - 4 b + 4 \cdot 8 + 2 b + 1$

$a = - b + 8$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.4.2.1.1.3

Multiply $4$ by $8$ .

$81 = - 4 b + 32 + 2 b + 1$

$a = - b + 8$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = - 4 b + 32 + 2 b + 1$

$a = - b + 8$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.4.2.1.2

Simplify by adding terms.

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Step 2.3.4.2.1.2.1

Add $- 4 b$ and $2 b$ .

$81 = - 2 b + 32 + 1$

$a = - b + 8$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.4.2.1.2.2

Add $32$ and $1$ .

$81 = - 2 b + 33$

$a = - b + 8$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = - 2 b + 33$

$a = - b + 8$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = - 2 b + 33$

$a = - b + 8$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$81 = - 2 b + 33$

$a = - b + 8$

$\frac{1}{9} = a - b + 1$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.4.3

Replace all occurrences of $a$ in $\frac{1}{9} = a - b + 1$ with $- b + 8$ .

$\frac{1}{9} = (- b + 8) - b + 1$

$81 = - 2 b + 33$

$a = - b + 8$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.4.4

Simplify the right side.

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Step 2.3.4.4.1

Simplify $(- b + 8) - b + 1$ .

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Step 2.3.4.4.1.1

Subtract $b$ from $- b$ .

$\frac{1}{9} = - 2 b + 8 + 1$

$81 = - 2 b + 33$

$a = - b + 8$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.4.4.1.2

Add $8$ and $1$ .

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$\frac{1}{81} = 4 a - 2 b + 1$

$c = 1$

Step 2.3.4.5

Replace all occurrences of $a$ in $\frac{1}{81} = 4 a - 2 b + 1$ with $- b + 8$ .

$\frac{1}{81} = 4 (- b + 8) - 2 b + 1$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.4.6

Simplify the right side.

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Step 2.3.4.6.1

Simplify $4 (- b + 8) - 2 b + 1$ .

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Step 2.3.4.6.1.1

Simplify each term.

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Step 2.3.4.6.1.1.1

Apply the distributive property.

$\frac{1}{81} = 4 (- b) + 4 \cdot 8 - 2 b + 1$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.4.6.1.1.2

Multiply $- 1$ by $4$ .

$\frac{1}{81} = - 4 b + 4 \cdot 8 - 2 b + 1$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.4.6.1.1.3

Multiply $4$ by $8$ .

$\frac{1}{81} = - 4 b + 32 - 2 b + 1$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$\frac{1}{81} = - 4 b + 32 - 2 b + 1$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.4.6.1.2

Simplify by adding terms.

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Step 2.3.4.6.1.2.1

Subtract $2 b$ from $- 4 b$ .

$\frac{1}{81} = - 6 b + 32 + 1$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.4.6.1.2.2

Add $32$ and $1$ .

$\frac{1}{81} = - 6 b + 33$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$\frac{1}{81} = - 6 b + 33$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$\frac{1}{81} = - 6 b + 33$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$\frac{1}{81} = - 6 b + 33$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$\frac{1}{81} = - 6 b + 33$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5

Solve for $b$ in $\frac{1}{81} = - 6 b + 33$ .

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Step 2.3.5.1

Rewrite the equation as $- 6 b + 33 = \frac{1}{81}$ .

$- 6 b + 33 = \frac{1}{81}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.2

Move all terms not containing $b$ to the right side of the equation.

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Step 2.3.5.2.1

Subtract $33$ from both sides of the equation.

$- 6 b = \frac{1}{81} - 33$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.2.2

To write $- 33$ as a fraction with a common denominator, multiply by $\frac{81}{81}$ .

$- 6 b = \frac{1}{81} - 33 \cdot \frac{81}{81}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.2.3

Combine $- 33$ and $\frac{81}{81}$ .

$- 6 b = \frac{1}{81} + \frac{- 33 \cdot 81}{81}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.2.4

Combine the numerators over the common denominator.

$- 6 b = \frac{1 - 33 \cdot 81}{81}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.2.5

Simplify the numerator.

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Step 2.3.5.2.5.1

Multiply $- 33$ by $81$ .

$- 6 b = \frac{1 - 2673}{81}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.2.5.2

Subtract $2673$ from $1$ .

$- 6 b = \frac{- 2672}{81}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$- 6 b = \frac{- 2672}{81}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.2.6

Move the negative in front of the fraction.

$- 6 b = - \frac{2672}{81}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$- 6 b = - \frac{2672}{81}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3

Divide each term in $- 6 b = - \frac{2672}{81}$ by $- 6$ and simplify.

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Step 2.3.5.3.1

Divide each term in $- 6 b = - \frac{2672}{81}$ by $- 6$ .

$\frac{- 6 b}{- 6} = \frac{- \frac{2672}{81}}{- 6}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3.2

Simplify the left side.

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Step 2.3.5.3.2.1

Cancel the common factor of $- 6$ .

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Step 2.3.5.3.2.1.1

Cancel the common factor.

$\frac{- 6 b}{- 6} = \frac{- \frac{2672}{81}}{- 6}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{- \frac{2672}{81}}{- 6}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$b = \frac{- \frac{2672}{81}}{- 6}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$b = \frac{- \frac{2672}{81}}{- 6}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3.3

Simplify the right side.

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Step 2.3.5.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$b = - \frac{2672}{81} \cdot \frac{1}{- 6}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3.3.2

Cancel the common factor of $2$ .

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Step 2.3.5.3.3.2.1

Move the leading negative in $- \frac{2672}{81}$ into the numerator.

$b = \frac{- 2672}{81} \cdot \frac{1}{- 6}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3.3.2.2

Factor $2$ out of $- 2672$ .

$b = \frac{2 (- 1336)}{81} \cdot \frac{1}{- 6}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3.3.2.3

Factor $2$ out of $- 6$ .

$b = \frac{2 \cdot - 1336}{81} \cdot \frac{1}{2 \cdot - 3}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3.3.2.4

Cancel the common factor.

$b = \frac{2 \cdot - 1336}{81} \cdot \frac{1}{2 \cdot - 3}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3.3.2.5

Rewrite the expression.

$b = \frac{- 1336}{81} \cdot \frac{1}{- 3}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$b = \frac{- 1336}{81} \cdot \frac{1}{- 3}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3.3.3

Multiply $\frac{- 1336}{81}$ by $\frac{1}{- 3}$ .

$b = \frac{- 1336}{81 \cdot - 3}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3.3.4

Multiply $81$ by $- 3$ .

$b = \frac{- 1336}{- 243}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.5.3.3.5

Dividing two negative values results in a positive value.

$b = \frac{1336}{243}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$b = \frac{1336}{243}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$b = \frac{1336}{243}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$b = \frac{1336}{243}$

$\frac{1}{9} = - 2 b + 9$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.6

Replace all occurrences of $b$ with $\frac{1336}{243}$ in each equation.

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Step 2.3.6.1

Replace all occurrences of $b$ in $\frac{1}{9} = - 2 b + 9$ with $\frac{1336}{243}$ .

$\frac{1}{9} = - 2 (\frac{1336}{243}) + 9$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.6.2

Simplify the right side.

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Step 2.3.6.2.1

Simplify $- 2 (\frac{1336}{243}) + 9$ .

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Step 2.3.6.2.1.1

Simplify each term.

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Step 2.3.6.2.1.1.1

Multiply $- 2 (\frac{1336}{243})$ .

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Step 2.3.6.2.1.1.1.1

Combine $- 2$ and $\frac{1336}{243}$ .

$\frac{1}{9} = \frac{- 2 \cdot 1336}{243} + 9$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.6.2.1.1.1.2

Multiply $- 2$ by $1336$ .

$\frac{1}{9} = \frac{- 2672}{243} + 9$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$\frac{1}{9} = \frac{- 2672}{243} + 9$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.6.2.1.1.2

Move the negative in front of the fraction.

$\frac{1}{9} = - \frac{2672}{243} + 9$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$\frac{1}{9} = - \frac{2672}{243} + 9$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.6.2.1.2

To write $9$ as a fraction with a common denominator, multiply by $\frac{243}{243}$ .

$\frac{1}{9} = - \frac{2672}{243} + 9 \cdot \frac{243}{243}$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.6.2.1.3

Combine $9$ and $\frac{243}{243}$ .

$\frac{1}{9} = - \frac{2672}{243} + \frac{9 \cdot 243}{243}$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.6.2.1.4

Combine the numerators over the common denominator.

$\frac{1}{9} = \frac{- 2672 + 9 \cdot 243}{243}$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.6.2.1.5

Simplify the numerator.

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Step 2.3.6.2.1.5.1

Multiply $9$ by $243$ .

$\frac{1}{9} = \frac{- 2672 + 2187}{243}$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.6.2.1.5.2

Add $- 2672$ and $2187$ .

$\frac{1}{9} = \frac{- 485}{243}$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$\frac{1}{9} = \frac{- 485}{243}$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.6.2.1.6

Move the negative in front of the fraction.

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$81 = - 2 b + 33$

$a = - b + 8$

$c = 1$

Step 2.3.6.3

Replace all occurrences of $b$ in $81 = - 2 b + 33$ with $\frac{1336}{243}$ .

$81 = - 2 (\frac{1336}{243}) + 33$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

Step 2.3.6.4

Simplify the right side.

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Step 2.3.6.4.1

Simplify $- 2 (\frac{1336}{243}) + 33$ .

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Step 2.3.6.4.1.1

Simplify each term.

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Step 2.3.6.4.1.1.1

Multiply $- 2 (\frac{1336}{243})$ .

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Step 2.3.6.4.1.1.1.1

Combine $- 2$ and $\frac{1336}{243}$ .

$81 = \frac{- 2 \cdot 1336}{243} + 33$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

Step 2.3.6.4.1.1.1.2

Multiply $- 2$ by $1336$ .

$81 = \frac{- 2672}{243} + 33$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

$81 = \frac{- 2672}{243} + 33$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

Step 2.3.6.4.1.1.2

Move the negative in front of the fraction.

$81 = - \frac{2672}{243} + 33$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

$81 = - \frac{2672}{243} + 33$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

Step 2.3.6.4.1.2

To write $33$ as a fraction with a common denominator, multiply by $\frac{243}{243}$ .

$81 = - \frac{2672}{243} + 33 \cdot \frac{243}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

Step 2.3.6.4.1.3

Combine $33$ and $\frac{243}{243}$ .

$81 = - \frac{2672}{243} + \frac{33 \cdot 243}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

Step 2.3.6.4.1.4

Combine the numerators over the common denominator.

$81 = \frac{- 2672 + 33 \cdot 243}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

Step 2.3.6.4.1.5

Simplify the numerator.

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Step 2.3.6.4.1.5.1

Multiply $33$ by $243$ .

$81 = \frac{- 2672 + 8019}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

Step 2.3.6.4.1.5.2

Add $- 2672$ and $8019$ .

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$a = - b + 8$

$c = 1$

Step 2.3.6.5

Replace all occurrences of $b$ in $a = - b + 8$ with $\frac{1336}{243}$ .

$a = - (\frac{1336}{243}) + 8$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$c = 1$

Step 2.3.6.6

Simplify the right side.

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Step 2.3.6.6.1

Simplify $- (\frac{1336}{243}) + 8$ .

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Step 2.3.6.6.1.1

To write $8$ as a fraction with a common denominator, multiply by $\frac{243}{243}$ .

$a = - \frac{1336}{243} + 8 \cdot \frac{243}{243}$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$c = 1$

Step 2.3.6.6.1.2

Combine $8$ and $\frac{243}{243}$ .

$a = - \frac{1336}{243} + \frac{8 \cdot 243}{243}$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$c = 1$

Step 2.3.6.6.1.3

Combine the numerators over the common denominator.

$a = \frac{- 1336 + 8 \cdot 243}{243}$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$c = 1$

Step 2.3.6.6.1.4

Simplify the numerator.

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Step 2.3.6.6.1.4.1

Multiply $8$ by $243$ .

$a = \frac{- 1336 + 1944}{243}$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$c = 1$

Step 2.3.6.6.1.4.2

Add $- 1336$ and $1944$ .

$a = \frac{608}{243}$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$c = 1$

$a = \frac{608}{243}$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$c = 1$

$a = \frac{608}{243}$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$c = 1$

$a = \frac{608}{243}$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$c = 1$

$a = \frac{608}{243}$

$81 = \frac{5347}{243}$

$\frac{1}{9} = - \frac{485}{243}$

$b = \frac{1336}{243}$

$c = 1$

Step 2.3.7

Since $81 = \frac{5347}{243}$ is not true, there is no solution.

No solution

Step 2.4

Calculate the value of $y$ using each $x$ value in the table and compare this value to the given $y$ value in the table.

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Step 2.4.1

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 0$ , $b = 0$ , $c = 0$ , and $x = - 2$ .

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Step 2.4.1.1

Simplify each term.

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Step 2.4.1.1.1

Raise $- 2$ to the power of $2$ .

$y = 0 \cdot 4 + (0) \cdot (- 2) + 0$

Step 2.4.1.1.2

Multiply $0$ by $4$ .

$y = 0 + (0) \cdot (- 2) + 0$

Step 2.4.1.1.3

Multiply $0$ by $- 2$ .

$y = 0 + 0 + 0$

Step 2.4.1.2

Simplify by adding numbers.

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Step 2.4.1.2.1

Add $0$ and $0$ .

$y = 0 + 0$

Step 2.4.1.2.2

Add $0$ and $0$ .

$y = 0$

Step 2.4.2

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = - 2$ . This check does not pass, since $y = 0$ and $y = \frac{1}{81}$ . The function rule can't be quadratic.

$0 \neq \frac{1}{81}$

Step 2.4.3

Since $y \neq y$ for the corresponding $x$ values, the function is not quadratic.

The function is not quadratic

Step 3

Check if the function rule is cubic.

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Step 3.1

To find if the table follows a function rule, check whether the function rule could follow the form $y = a x^{3} + b x^{2} + c x + d$ .

$y = a x^{3} + b x^{2} + c x + d$

Step 3.2

Build a set of $4$ equations from the table such that $y = a x^{3} + b x^{2} + c x + d$ .

Step 3.3

Calculate the values of $a$ , $b$ , $c$ , and $d$ .

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Step 3.3.1

Solve for $d$ in $\frac{1}{81} = a {(- 2)}^{3} + b {(- 2)}^{2} + c (- 2) + d$ .

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Step 3.3.1.1

Rewrite the equation as $a {(- 2)}^{3} + b {(- 2)}^{2} + c (- 2) + d = \frac{1}{81}$ .

$a {(- 2)}^{3} + b {(- 2)}^{2} + c (- 2) + d = \frac{1}{81}$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.1.2

Simplify each term.

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Step 3.3.1.2.1

Raise $- 2$ to the power of $3$ .

$a \cdot - 8 + b {(- 2)}^{2} + c (- 2) + d = \frac{1}{81}$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.1.2.2

Move $- 8$ to the left of $a$ .

$- 8 \cdot a + b {(- 2)}^{2} + c (- 2) + d = \frac{1}{81}$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.1.2.3

Raise $- 2$ to the power of $2$ .

$- 8 a + b \cdot 4 + c (- 2) + d = \frac{1}{81}$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.1.2.4

Move $4$ to the left of $b$ .

$- 8 a + 4 \cdot b + c (- 2) + d = \frac{1}{81}$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.1.2.5

Move $- 2$ to the left of $c$ .

$- 8 a + 4 b - 2 c + d = \frac{1}{81}$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$- 8 a + 4 b - 2 c + d = \frac{1}{81}$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.1.3

Move all terms not containing $d$ to the right side of the equation.

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Step 3.3.1.3.1

Add $8 a$ to both sides of the equation.

$4 b - 2 c + d = \frac{1}{81} + 8 a$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.1.3.2

Subtract $4 b$ from both sides of the equation.

$- 2 c + d = \frac{1}{81} + 8 a - 4 b$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.1.3.3

Add $2 c$ to both sides of the equation.

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2

Replace all occurrences of $d$ with $\frac{1}{81} + 8 a - 4 b + 2 c$ in each equation.

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Step 3.3.2.1

Replace all occurrences of $d$ in $\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + d$ with $\frac{1}{81} + 8 a - 4 b + 2 c$ .

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2

Simplify $\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.2.1

Simplify the left side.

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Step 3.3.2.2.1.1

Remove parentheses.

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2.2

Simplify the right side.

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Step 3.3.2.2.2.1

Simplify $a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.2.2.1.1

Combine the opposite terms in $a {(- 1)}^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.2.2.1.1.1

Reorder the factors in the terms $c (- 2)$ and $2 c$ .

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} - 2 c + \frac{1}{81} + 8 a - 4 b + 2 c$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2.2.1.1.2

Add $- 2 c$ and $2 c$ .

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b + 0$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2.2.1.1.3

Add $a {(- 1)}^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$ and $0$ .

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$\frac{1}{9} = a {(- 1)}^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2.2.1.2

Simplify each term.

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Step 3.3.2.2.2.1.2.1

Raise $- 1$ to the power of $3$ .

$\frac{1}{9} = a \cdot - 1 + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2.2.1.2.2

Move $- 1$ to the left of $a$ .

$\frac{1}{9} = - 1 \cdot a + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2.2.1.2.3

Rewrite $- 1 a$ as $- a$ .

$\frac{1}{9} = - a + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2.2.1.2.4

Raise $- 2$ to the power of $2$ .

$\frac{1}{9} = - a + b \cdot 4 + \frac{1}{81} + 8 a - 4 b$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2.2.1.2.5

Move $4$ to the left of $b$ .

$\frac{1}{9} = - a + 4 b + \frac{1}{81} + 8 a - 4 b$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$\frac{1}{9} = - a + 4 b + \frac{1}{81} + 8 a - 4 b$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2.2.1.3

Simplify by adding terms.

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Step 3.3.2.2.2.1.3.1

Combine the opposite terms in $- a + 4 b + \frac{1}{81} + 8 a - 4 b$ .

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Step 3.3.2.2.2.1.3.1.1

Subtract $4 b$ from $4 b$ .

$\frac{1}{9} = - a + \frac{1}{81} + 8 a + 0$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2.2.1.3.1.2

Add $- a + \frac{1}{81} + 8 a$ and $0$ .

$\frac{1}{9} = - a + \frac{1}{81} + 8 a$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$\frac{1}{9} = - a + \frac{1}{81} + 8 a$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.2.2.1.3.2

Add $- a$ and $8 a$ .

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.3

Replace all occurrences of $d$ in $1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + d$ with $\frac{1}{81} + 8 a - 4 b + 2 c$ .

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.4

Simplify $1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.4.1

Simplify the left side.

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Step 3.3.2.4.1.1

Remove parentheses.

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.4.2

Simplify the right side.

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Step 3.3.2.4.2.1

Simplify $a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.4.2.1.1

Combine the opposite terms in $a \cdot 0^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.4.2.1.1.1

Reorder the factors in the terms $c (- 2)$ and $2 c$ .

$1 = a \cdot 0^{3} + b {(- 2)}^{2} - 2 c + \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.4.2.1.1.2

Add $- 2 c$ and $2 c$ .

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b + 0$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.4.2.1.1.3

Add $a \cdot 0^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$ and $0$ .

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = a \cdot 0^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.4.2.1.2

Simplify each term.

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Step 3.3.2.4.2.1.2.1

Raising $0$ to any positive power yields $0$ .

$1 = a \cdot 0 + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.4.2.1.2.2

Multiply $a$ by $0$ .

$1 = 0 + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.4.2.1.2.3

Raise $- 2$ to the power of $2$ .

$1 = 0 + b \cdot 4 + \frac{1}{81} + 8 a - 4 b$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.4.2.1.2.4

Move $4$ to the left of $b$ .

$1 = 0 + 4 b + \frac{1}{81} + 8 a - 4 b$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = 0 + 4 b + \frac{1}{81} + 8 a - 4 b$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.4.2.1.3

Combine the opposite terms in $0 + 4 b + \frac{1}{81} + 8 a - 4 b$ .

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Step 3.3.2.4.2.1.3.1

Add $0$ and $4 b$ .

$1 = 4 b + \frac{1}{81} + 8 a - 4 b$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.4.2.1.3.2

Subtract $4 b$ from $4 b$ .

$1 = \frac{1}{81} + 8 a + 0$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.4.2.1.3.3

Add $\frac{1}{81} + 8 a$ and $0$ .

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = a + b {(- 2)}^{2} + c (- 2) + d$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.5

Replace all occurrences of $d$ in $9 = a + b {(- 2)}^{2} + c (- 2) + d$ with $\frac{1}{81} + 8 a - 4 b + 2 c$ .

$9 = a + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.6

Simplify $9 = a + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.6.1

Simplify the left side.

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Step 3.3.2.6.1.1

Remove parentheses.

$9 = a + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.6.2

Simplify the right side.

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Step 3.3.2.6.2.1

Simplify $a + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.6.2.1.1

Combine the opposite terms in $a + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.6.2.1.1.1

Reorder the factors in the terms $c (- 2)$ and $2 c$ .

$9 = a + b {(- 2)}^{2} - 2 c + \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.6.2.1.1.2

Add $- 2 c$ and $2 c$ .

$9 = a + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b + 0$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.6.2.1.1.3

Add $a + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$ and $0$ .

$9 = a + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.6.2.1.2

Simplify each term.

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Step 3.3.2.6.2.1.2.1

Raise $- 2$ to the power of $2$ .

$9 = a + b \cdot 4 + \frac{1}{81} + 8 a - 4 b$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.6.2.1.2.2

Move $4$ to the left of $b$ .

$9 = a + 4 b + \frac{1}{81} + 8 a - 4 b$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + 4 b + \frac{1}{81} + 8 a - 4 b$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.6.2.1.3

Simplify by adding terms.

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Step 3.3.2.6.2.1.3.1

Combine the opposite terms in $a + 4 b + \frac{1}{81} + 8 a - 4 b$ .

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Step 3.3.2.6.2.1.3.1.1

Subtract $4 b$ from $4 b$ .

$9 = a + \frac{1}{81} + 8 a + 0$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.6.2.1.3.1.2

Add $a + \frac{1}{81} + 8 a$ and $0$ .

$9 = a + \frac{1}{81} + 8 a$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = a + \frac{1}{81} + 8 a$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.6.2.1.3.2

Add $a$ and $8 a$ .

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$

Step 3.3.2.7

Replace all occurrences of $d$ in $81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + d$ with $\frac{1}{81} + 8 a - 4 b + 2 c$ .

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.2.8

Simplify $81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.8.1

Simplify the left side.

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Step 3.3.2.8.1.1

Remove parentheses.

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.2.8.2

Simplify the right side.

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Step 3.3.2.8.2.1

Simplify $a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.8.2.1.1

Combine the opposite terms in $a \cdot 2^{3} + b {(- 2)}^{2} + c (- 2) + \frac{1}{81} + 8 a - 4 b + 2 c$ .

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Step 3.3.2.8.2.1.1.1

Reorder the factors in the terms $c (- 2)$ and $2 c$ .

$81 = a \cdot 2^{3} + b {(- 2)}^{2} - 2 c + \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.2.8.2.1.1.2

Add $- 2 c$ and $2 c$ .

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b + 0$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.2.8.2.1.1.3

Add $a \cdot 2^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$ and $0$ .

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = a \cdot 2^{3} + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.2.8.2.1.2

Simplify each term.

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Step 3.3.2.8.2.1.2.1

Raise $2$ to the power of $3$ .

$81 = a \cdot 8 + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.2.8.2.1.2.2

Move $8$ to the left of $a$ .

$81 = 8 \cdot a + b {(- 2)}^{2} + \frac{1}{81} + 8 a - 4 b$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.2.8.2.1.2.3

Raise $- 2$ to the power of $2$ .

$81 = 8 a + b \cdot 4 + \frac{1}{81} + 8 a - 4 b$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.2.8.2.1.2.4

Move $4$ to the left of $b$ .

$81 = 8 a + 4 b + \frac{1}{81} + 8 a - 4 b$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = 8 a + 4 b + \frac{1}{81} + 8 a - 4 b$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.2.8.2.1.3

Simplify by adding terms.

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Step 3.3.2.8.2.1.3.1

Combine the opposite terms in $8 a + 4 b + \frac{1}{81} + 8 a - 4 b$ .

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Step 3.3.2.8.2.1.3.1.1

Subtract $4 b$ from $4 b$ .

$81 = 8 a + \frac{1}{81} + 8 a + 0$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.2.8.2.1.3.1.2

Add $8 a + \frac{1}{81} + 8 a$ and $0$ .

$81 = 8 a + \frac{1}{81} + 8 a$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = 8 a + \frac{1}{81} + 8 a$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.2.8.2.1.3.2

Add $8 a$ and $8 a$ .

$81 = 16 a + \frac{1}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = 16 a + \frac{1}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = 16 a + \frac{1}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = 16 a + \frac{1}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = 16 a + \frac{1}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$81 = 16 a + \frac{1}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3

Solve for $a$ in $81 = 16 a + \frac{1}{81}$ .

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Step 3.3.3.1

Rewrite the equation as $16 a + \frac{1}{81} = 81$ .

$16 a + \frac{1}{81} = 81$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.2

Move all terms not containing $a$ to the right side of the equation.

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Step 3.3.3.2.1

Subtract $\frac{1}{81}$ from both sides of the equation.

$16 a = 81 - \frac{1}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.2.2

To write $81$ as a fraction with a common denominator, multiply by $\frac{81}{81}$ .

$16 a = 81 \cdot \frac{81}{81} - \frac{1}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.2.3

Combine $81$ and $\frac{81}{81}$ .

$16 a = \frac{81 \cdot 81}{81} - \frac{1}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.2.4

Combine the numerators over the common denominator.

$16 a = \frac{81 \cdot 81 - 1}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.2.5

Simplify the numerator.

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Step 3.3.3.2.5.1

Multiply $81$ by $81$ .

$16 a = \frac{6561 - 1}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.2.5.2

Subtract $1$ from $6561$ .

$16 a = \frac{6560}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$16 a = \frac{6560}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$16 a = \frac{6560}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.3

Divide each term in $16 a = \frac{6560}{81}$ by $16$ and simplify.

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Step 3.3.3.3.1

Divide each term in $16 a = \frac{6560}{81}$ by $16$ .

$\frac{16 a}{16} = \frac{\frac{6560}{81}}{16}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.3.2

Simplify the left side.

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Step 3.3.3.3.2.1

Cancel the common factor of $16$ .

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Step 3.3.3.3.2.1.1

Cancel the common factor.

$\frac{16 a}{16} = \frac{\frac{6560}{81}}{16}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{\frac{6560}{81}}{16}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$a = \frac{\frac{6560}{81}}{16}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$a = \frac{\frac{6560}{81}}{16}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.3.3

Simplify the right side.

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Step 3.3.3.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$a = \frac{6560}{81} \cdot \frac{1}{16}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.3.3.2

Cancel the common factor of $16$ .

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Step 3.3.3.3.3.2.1

Factor $16$ out of $6560$ .

$a = \frac{16 (410)}{81} \cdot \frac{1}{16}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.3.3.2.2

Cancel the common factor.

$a = \frac{16 \cdot 410}{81} \cdot \frac{1}{16}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.3.3.3.2.3

Rewrite the expression.

$a = \frac{410}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$a = \frac{410}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$a = \frac{410}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$a = \frac{410}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$a = \frac{410}{81}$

$9 = 9 a + \frac{1}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4

Replace all occurrences of $a$ with $\frac{410}{81}$ in each equation.

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Step 3.3.4.1

Replace all occurrences of $a$ in $9 = 9 a + \frac{1}{81}$ with $\frac{410}{81}$ .

$9 = 9 (\frac{410}{81}) + \frac{1}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.2

Simplify the right side.

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Step 3.3.4.2.1

Simplify $9 (\frac{410}{81}) + \frac{1}{81}$ .

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Step 3.3.4.2.1.1

Cancel the common factor of $9$ .

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Step 3.3.4.2.1.1.1

Factor $9$ out of $81$ .

$9 = 9 (\frac{410}{9 (9)}) + \frac{1}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.2.1.1.2

Cancel the common factor.

$9 = 9 (\frac{410}{9 \cdot 9}) + \frac{1}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.2.1.1.3

Rewrite the expression.

$9 = \frac{410}{9} + \frac{1}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = \frac{410}{9} + \frac{1}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.2.1.2

To write $\frac{410}{9}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$9 = \frac{410}{9} \cdot \frac{9}{9} + \frac{1}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.2.1.3

Write each expression with a common denominator of $81$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.3.4.2.1.3.1

Multiply $\frac{410}{9}$ by $\frac{9}{9}$ .

$9 = \frac{410 \cdot 9}{9 \cdot 9} + \frac{1}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.2.1.3.2

Multiply $9$ by $9$ .

$9 = \frac{410 \cdot 9}{81} + \frac{1}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = \frac{410 \cdot 9}{81} + \frac{1}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.2.1.4

Combine the numerators over the common denominator.

$9 = \frac{410 \cdot 9 + 1}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.2.1.5

Simplify the numerator.

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Step 3.3.4.2.1.5.1

Multiply $410$ by $9$ .

$9 = \frac{3690 + 1}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.2.1.5.2

Add $3690$ and $1$ .

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$1 = \frac{1}{81} + 8 a$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.3

Replace all occurrences of $a$ in $1 = \frac{1}{81} + 8 a$ with $\frac{410}{81}$ .

$1 = \frac{1}{81} + 8 (\frac{410}{81})$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.4

Simplify the right side.

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Step 3.3.4.4.1

Simplify $\frac{1}{81} + 8 (\frac{410}{81})$ .

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Step 3.3.4.4.1.1

Multiply $8 (\frac{410}{81})$ .

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Step 3.3.4.4.1.1.1

Combine $8$ and $\frac{410}{81}$ .

$1 = \frac{1}{81} + \frac{8 \cdot 410}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.4.1.1.2

Multiply $8$ by $410$ .

$1 = \frac{1}{81} + \frac{3280}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = \frac{1}{81} + \frac{3280}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.4.1.2

Combine the numerators over the common denominator.

$1 = \frac{1 + 3280}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.4.1.3

Add $1$ and $3280$ .

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$\frac{1}{9} = 7 a + \frac{1}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.5

Replace all occurrences of $a$ in $\frac{1}{9} = 7 a + \frac{1}{81}$ with $\frac{410}{81}$ .

$\frac{1}{9} = 7 (\frac{410}{81}) + \frac{1}{81}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.6

Simplify the right side.

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Step 3.3.4.6.1

Simplify $7 (\frac{410}{81}) + \frac{1}{81}$ .

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Step 3.3.4.6.1.1

Multiply $7 (\frac{410}{81})$ .

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Step 3.3.4.6.1.1.1

Combine $7$ and $\frac{410}{81}$ .

$\frac{1}{9} = \frac{7 \cdot 410}{81} + \frac{1}{81}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.6.1.1.2

Multiply $7$ by $410$ .

$\frac{1}{9} = \frac{2870}{81} + \frac{1}{81}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = \frac{2870}{81} + \frac{1}{81}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.6.1.2

Combine the numerators over the common denominator.

$\frac{1}{9} = \frac{2870 + 1}{81}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.6.1.3

Add $2870$ and $1$ .

$\frac{1}{9} = \frac{2871}{81}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.6.1.4

Cancel the common factor of $2871$ and $81$ .

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Step 3.3.4.6.1.4.1

Factor $9$ out of $2871$ .

$\frac{1}{9} = \frac{9 (319)}{81}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.6.1.4.2

Cancel the common factors.

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Step 3.3.4.6.1.4.2.1

Factor $9$ out of $81$ .

$\frac{1}{9} = \frac{9 \cdot 319}{9 \cdot 9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.6.1.4.2.2

Cancel the common factor.

$\frac{1}{9} = \frac{9 \cdot 319}{9 \cdot 9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.6.1.4.2.3

Rewrite the expression.

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + 8 a - 4 b + 2 c$

Step 3.3.4.7

Replace all occurrences of $a$ in $d = \frac{1}{81} + 8 a - 4 b + 2 c$ with $\frac{410}{81}$ .

$d = \frac{1}{81} + 8 (\frac{410}{81}) - 4 b + 2 c$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

Step 3.3.4.8

Simplify the right side.

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Step 3.3.4.8.1

Simplify $\frac{1}{81} + 8 (\frac{410}{81}) - 4 b + 2 c$ .

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Step 3.3.4.8.1.1

Multiply $8 (\frac{410}{81})$ .

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Step 3.3.4.8.1.1.1

Combine $8$ and $\frac{410}{81}$ .

$d = \frac{1}{81} + \frac{8 \cdot 410}{81} - 4 b + 2 c$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

Step 3.3.4.8.1.1.2

Multiply $8$ by $410$ .

$d = \frac{1}{81} + \frac{3280}{81} - 4 b + 2 c$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = \frac{1}{81} + \frac{3280}{81} - 4 b + 2 c$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

Step 3.3.4.8.1.2

Combine the numerators over the common denominator.

$d = - 4 b + 2 c + \frac{1 + 3280}{81}$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

Step 3.3.4.8.1.3

Add $1$ and $3280$ .

$d = - 4 b + 2 c + \frac{3281}{81}$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = - 4 b + 2 c + \frac{3281}{81}$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = - 4 b + 2 c + \frac{3281}{81}$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

$d = - 4 b + 2 c + \frac{3281}{81}$

$\frac{1}{9} = \frac{319}{9}$

$1 = \frac{3281}{81}$

$9 = \frac{3691}{81}$

$a = \frac{410}{81}$

Step 3.3.5

Since $\frac{1}{9} = \frac{319}{9}$ is not true, there is no solution.

No solution

Step 3.4

Calculate the value of $y$ using each $x$ value in the table and compare this value to the given $y$ value in the table.

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Step 3.4.1

Calculate the value of $y$ such that $y = a x^{3} + b$ when $a = 0$ , $b = 0$ , $c = 0$ , $d = 0$ , and $x = - 2$ .

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Step 3.4.1.1

Simplify each term.

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Step 3.4.1.1.1

Raise $- 2$ to the power of $3$ .

$y = 0 \cdot - 8 + (0) \cdot (- 2^{2}) + (0) \cdot ((- 2) \cdot 0)$

Step 3.4.1.1.2

Multiply $0$ by $- 8$ .

$y = 0 + (0) \cdot (- 2^{2}) + (0) \cdot ((- 2) \cdot 0)$

Step 3.4.1.1.3

Raise $2$ to the power of $2$ .

$y = 0 + 0 \cdot (- 1 \cdot 4) + (0) \cdot ((- 2) \cdot 0)$

Step 3.4.1.1.4

Multiply $0 (- 1 \cdot 4)$ .

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Step 3.4.1.1.4.1

Multiply $- 1$ by $4$ .

$y = 0 + 0 \cdot - 4 + (0) \cdot ((- 2) \cdot 0)$

Step 3.4.1.1.4.2

Multiply $0$ by $- 4$ .

$y = 0 + 0 + (0) \cdot ((- 2) \cdot 0)$

Step 3.4.1.1.5

Multiply $(0) ((- 2) \cdot 0)$ .

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Step 3.4.1.1.5.1

Multiply $- 2$ by $0$ .

$y = 0 + 0 + 0 \cdot 0$

Step 3.4.1.1.5.2

Multiply $0$ by $0$ .

$y = 0 + 0 + 0$

Step 3.4.1.2

Simplify by adding numbers.

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Step 3.4.1.2.1

Add $0$ and $0$ .

$y = 0 + 0$

Step 3.4.1.2.2

Add $0$ and $0$ .

$y = 0$

Step 3.4.2

If the table has a cubic function rule, $y = y$ for the corresponding $x$ value, $x = - 2$ . This check does not pass, since $y = 0$ and $y = \frac{1}{81}$ . The function rule can't be cubic.

$0 \neq \frac{1}{81}$

Step 3.4.3

Since $y \neq y$ for the corresponding $x$ values, the function is not cubic.

The function is not cubic

Step 4

There are no values for $a$ , $b$ , $c$ , and $d$ in the equations $y = a x + b$ , $y = a x^{2} + b x + c$ , and $y = a x^{3} + b x^{2} + c x + d$ that work for every pair of $x$ and $y$ .

The table does not have a function rule that is linear, quadratic, or cubic.