Pre-Algebra Examples

$\begin{array}{c} x & y \\ - 5 & 10 \\ 7 & - 2 \\ 0 & 15 \\ 4 & - 5 \end{array}$

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 $y = a x + b$ .

$10 = a (- 5) + b - 2 = a (7) + b 15 = a (0) + b - 5 = a (4) + b$

Step 1.3

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

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Step 1.3.1

Rewrite the equation as $b = 15$ .

$b = 15$

$10 = a (- 5) + b$

$- 2 = a (7) + b$

$- 5 = a (4) + b$

Step 1.3.2

Replace all occurrences of $b$ with $15$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $b$ in $10 = a (- 5) + b$ with $15$ .

$10 = a (- 5) + 15$

$b = 15$

$- 2 = a (7) + b$

$- 5 = a (4) + b$

Step 1.3.2.2

Simplify $10 = a (- 5) + 15$ .

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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.

$10 = a (- 5) + 15$

$b = 15$

$- 2 = a (7) + b$

$- 5 = a (4) + b$

$10 = a (- 5) + 15$

$b = 15$

$- 2 = a (7) + b$

$- 5 = a (4) + b$

Step 1.3.2.2.2

Simplify the right side.

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Step 1.3.2.2.2.1

Move $- 5$ to the left of $a$ .

$10 = - 5 a + 15$

$b = 15$

$- 2 = a (7) + b$

$- 5 = a (4) + b$

$10 = - 5 a + 15$

$b = 15$

$- 2 = a (7) + b$

$- 5 = a (4) + b$

$10 = - 5 a + 15$

$b = 15$

$- 2 = a (7) + b$

$- 5 = a (4) + b$

Step 1.3.2.3

Replace all occurrences of $b$ in $- 2 = a (7) + b$ with $15$ .

$- 2 = a (7) + 15$

$10 = - 5 a + 15$

$b = 15$

$- 5 = a (4) + b$

Step 1.3.2.4

Simplify $- 2 = a (7) + 15$ .

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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.

$- 2 = a (7) + 15$

$10 = - 5 a + 15$

$b = 15$

$- 5 = a (4) + b$

$- 2 = a (7) + 15$

$10 = - 5 a + 15$

$b = 15$

$- 5 = a (4) + b$

Step 1.3.2.4.2

Simplify the right side.

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Step 1.3.2.4.2.1

Move $7$ to the left of $a$ .

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$- 5 = a (4) + b$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$- 5 = a (4) + b$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$- 5 = a (4) + b$

Step 1.3.2.5

Replace all occurrences of $b$ in $- 5 = a (4) + b$ with $15$ .

$- 5 = a (4) + 15$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

Step 1.3.2.6

Simplify $- 5 = a (4) + 15$ .

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Step 1.3.2.6.1

Simplify the left side.

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Step 1.3.2.6.1.1

Remove parentheses.

$- 5 = a (4) + 15$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$- 5 = a (4) + 15$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

Step 1.3.2.6.2

Simplify the right side.

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Step 1.3.2.6.2.1

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

$- 5 = 4 a + 15$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$- 5 = 4 a + 15$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$- 5 = 4 a + 15$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$- 5 = 4 a + 15$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

Step 1.3.3

Solve for $a$ in $- 5 = 4 a + 15$ .

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Step 1.3.3.1

Rewrite the equation as $4 a + 15 = - 5$ .

$4 a + 15 = - 5$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

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 $15$ from both sides of the equation.

$4 a = - 5 - 15$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

Step 1.3.3.2.2

Subtract $15$ from $- 5$ .

$4 a = - 20$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$4 a = - 20$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

Step 1.3.3.3

Divide each term in $4 a = - 20$ by $4$ and simplify.

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Step 1.3.3.3.1

Divide each term in $4 a = - 20$ by $4$ .

$\frac{4 a}{4} = \frac{- 20}{4}$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

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 $4$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{4 a}{4} = \frac{- 20}{4}$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

Step 1.3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{- 20}{4}$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$a = \frac{- 20}{4}$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$a = \frac{- 20}{4}$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide $- 20$ by $4$ .

$a = - 5$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$a = - 5$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$a = - 5$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

$a = - 5$

$- 2 = 7 a + 15$

$10 = - 5 a + 15$

$b = 15$

Step 1.3.4

Replace all occurrences of $a$ with $- 5$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $a$ in $- 2 = 7 a + 15$ with $- 5$ .

$- 2 = 7 (- 5) + 15$

$a = - 5$

$10 = - 5 a + 15$

$b = 15$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $7 (- 5) + 15$ .

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Step 1.3.4.2.1.1

Multiply $7$ by $- 5$ .

$- 2 = - 35 + 15$

$a = - 5$

$10 = - 5 a + 15$

$b = 15$

Step 1.3.4.2.1.2

Add $- 35$ and $15$ .

$- 2 = - 20$

$a = - 5$

$10 = - 5 a + 15$

$b = 15$

$- 2 = - 20$

$a = - 5$

$10 = - 5 a + 15$

$b = 15$

$- 2 = - 20$

$a = - 5$

$10 = - 5 a + 15$

$b = 15$

Step 1.3.4.3

Replace all occurrences of $a$ in $10 = - 5 a + 15$ with $- 5$ .

$10 = - 5 \cdot - 5 + 15$

$- 2 = - 20$

$a = - 5$

$b = 15$

Step 1.3.4.4

Simplify the right side.

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Step 1.3.4.4.1

Simplify $- 5 \cdot - 5 + 15$ .

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Step 1.3.4.4.1.1

Multiply $- 5$ by $- 5$ .

$10 = 25 + 15$

$- 2 = - 20$

$a = - 5$

$b = 15$

Step 1.3.4.4.1.2

Add $25$ and $15$ .

$10 = 40$

$- 2 = - 20$

$a = - 5$

$b = 15$

$10 = 40$

$- 2 = - 20$

$a = - 5$

$b = 15$

$10 = 40$

$- 2 = - 20$

$a = - 5$

$b = 15$

$10 = 40$

$- 2 = - 20$

$a = - 5$

$b = 15$

Step 1.3.5

Since $10 = 40$ 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 $15 = a \cdot 0^{2} + c$ .

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Step 2.3.1.1

Rewrite the equation as $a \cdot 0^{2} + c = 15$ .

$a \cdot 0^{2} + c = 15$

$10 = a {(- 5)}^{2} + b (- 5) + c$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

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 $0$ .

$a \cdot 0 + c = 15$

$10 = a {(- 5)}^{2} + b (- 5) + c$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

Step 2.3.1.2.1.2

Multiply $a$ by $0$ .

$0 + c = 15$

$10 = a {(- 5)}^{2} + b (- 5) + c$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

$0 + c = 15$

$10 = a {(- 5)}^{2} + b (- 5) + c$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

Step 2.3.1.2.2

Add $0$ and $c$ .

$c = 15$

$10 = a {(- 5)}^{2} + b (- 5) + c$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

$c = 15$

$10 = a {(- 5)}^{2} + b (- 5) + c$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

$c = 15$

$10 = a {(- 5)}^{2} + b (- 5) + c$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

Step 2.3.2

Replace all occurrences of $c$ with $15$ in each equation.

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Step 2.3.2.1

Replace all occurrences of $c$ in $10 = a {(- 5)}^{2} + b (- 5) + c$ with $15$ .

$10 = a {(- 5)}^{2} + b (- 5) + 15$

$c = 15$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

Step 2.3.2.2

Simplify $10 = a {(- 5)}^{2} + b (- 5) + 15$ .

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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.

$10 = a {(- 5)}^{2} + b (- 5) + 15$

$c = 15$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

$10 = a {(- 5)}^{2} + b (- 5) + 15$

$c = 15$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

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 $- 5$ to the power of $2$ .

$10 = a \cdot 25 + b (- 5) + 15$

$c = 15$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

Step 2.3.2.2.2.1.2

Move $25$ to the left of $a$ .

$10 = 25 \cdot a + b (- 5) + 15$

$c = 15$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

Step 2.3.2.2.2.1.3

Move $- 5$ to the left of $b$ .

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = a \cdot 7^{2} + b (7) + c$

$- 5 = a \cdot 4^{2} + b (4) + c$

Step 2.3.2.3

Replace all occurrences of $c$ in $- 2 = a \cdot 7^{2} + b (7) + c$ with $15$ .

$- 2 = a \cdot 7^{2} + b (7) + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = a \cdot 4^{2} + b (4) + c$

Step 2.3.2.4

Simplify $- 2 = a \cdot 7^{2} + b (7) + 15$ .

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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.

$- 2 = a \cdot 7^{2} + b (7) + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = a \cdot 4^{2} + b (4) + c$

$- 2 = a \cdot 7^{2} + b (7) + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = a \cdot 4^{2} + b (4) + c$

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 $7$ to the power of $2$ .

$- 2 = a \cdot 49 + b (7) + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = a \cdot 4^{2} + b (4) + c$

Step 2.3.2.4.2.1.2

Move $49$ to the left of $a$ .

$- 2 = 49 \cdot a + b (7) + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = a \cdot 4^{2} + b (4) + c$

Step 2.3.2.4.2.1.3

Move $7$ to the left of $b$ .

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = a \cdot 4^{2} + b (4) + c$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = a \cdot 4^{2} + b (4) + c$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = a \cdot 4^{2} + b (4) + c$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = a \cdot 4^{2} + b (4) + c$

Step 2.3.2.5

Replace all occurrences of $c$ in $- 5 = a \cdot 4^{2} + b (4) + c$ with $15$ .

$- 5 = a \cdot 4^{2} + b (4) + 15$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.2.6

Simplify $- 5 = a \cdot 4^{2} + b (4) + 15$ .

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Step 2.3.2.6.1

Simplify the left side.

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Step 2.3.2.6.1.1

Remove parentheses.

$- 5 = a \cdot 4^{2} + b (4) + 15$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = a \cdot 4^{2} + b (4) + 15$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.2.6.2

Simplify the right side.

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Step 2.3.2.6.2.1

Simplify each term.

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Step 2.3.2.6.2.1.1

Raise $4$ to the power of $2$ .

$- 5 = a \cdot 16 + b (4) + 15$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.2.6.2.1.2

Move $16$ to the left of $a$ .

$- 5 = 16 \cdot a + b (4) + 15$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.2.6.2.1.3

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

$- 5 = 16 a + 4 b + 15$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = 16 a + 4 b + 15$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = 16 a + 4 b + 15$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = 16 a + 4 b + 15$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 5 = 16 a + 4 b + 15$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3

Solve for $a$ in $- 5 = 16 a + 4 b + 15$ .

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Step 2.3.3.1

Rewrite the equation as $16 a + 4 b + 15 = - 5$ .

$16 a + 4 b + 15 = - 5$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

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 $4 b$ from both sides of the equation.

$16 a + 15 = - 5 - 4 b$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.2.2

Subtract $15$ from both sides of the equation.

$16 a = - 5 - 4 b - 15$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.2.3

Subtract $15$ from $- 5$ .

$16 a = - 4 b - 20$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$16 a = - 4 b - 20$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3

Divide each term in $16 a = - 4 b - 20$ by $16$ and simplify.

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Step 2.3.3.3.1

Divide each term in $16 a = - 4 b - 20$ by $16$ .

$\frac{16 a}{16} = \frac{- 4 b}{16} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.2

Simplify the left side.

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Step 2.3.3.3.2.1

Cancel the common factor of $16$ .

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Step 2.3.3.3.2.1.1

Cancel the common factor.

$\frac{16 a}{16} = \frac{- 4 b}{16} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{- 4 b}{16} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$a = \frac{- 4 b}{16} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$a = \frac{- 4 b}{16} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.3

Simplify the right side.

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Step 2.3.3.3.3.1

Simplify each term.

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Step 2.3.3.3.3.1.1

Cancel the common factor of $- 4$ and $16$ .

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Step 2.3.3.3.3.1.1.1

Factor $4$ out of $- 4 b$ .

$a = \frac{4 (- b)}{16} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.3.1.1.2

Cancel the common factors.

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Step 2.3.3.3.3.1.1.2.1

Factor $4$ out of $16$ .

$a = \frac{4 (- b)}{4 (4)} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.3.1.1.2.2

Cancel the common factor.

$a = \frac{4 (- b)}{4 \cdot 4} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.3.1.1.2.3

Rewrite the expression.

$a = \frac{- b}{4} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$a = \frac{- b}{4} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$a = \frac{- b}{4} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.3.1.2

Move the negative in front of the fraction.

$a = - \frac{b}{4} + \frac{- 20}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.3.1.3

Cancel the common factor of $- 20$ and $16$ .

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Step 2.3.3.3.3.1.3.1

Factor $4$ out of $- 20$ .

$a = - \frac{b}{4} + \frac{4 (- 5)}{16}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.3.1.3.2

Cancel the common factors.

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Step 2.3.3.3.3.1.3.2.1

Factor $4$ out of $16$ .

$a = - \frac{b}{4} + \frac{4 \cdot - 5}{4 \cdot 4}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.3.1.3.2.2

Cancel the common factor.

$a = - \frac{b}{4} + \frac{4 \cdot - 5}{4 \cdot 4}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.3.1.3.2.3

Rewrite the expression.

$a = - \frac{b}{4} + \frac{- 5}{4}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$a = - \frac{b}{4} + \frac{- 5}{4}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$a = - \frac{b}{4} + \frac{- 5}{4}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.3.3.3.1.4

Move the negative in front of the fraction.

$a = - \frac{b}{4} - \frac{5}{4}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$- 2 = 49 a + 7 b + 15$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4

Replace all occurrences of $a$ with $- \frac{b}{4} - \frac{5}{4}$ in each equation.

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Step 2.3.4.1

Replace all occurrences of $a$ in $- 2 = 49 a + 7 b + 15$ with $- \frac{b}{4} - \frac{5}{4}$ .

$- 2 = 49 (- \frac{b}{4} - \frac{5}{4}) + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2

Simplify the right side.

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Step 2.3.4.2.1

Simplify $49 (- \frac{b}{4} - \frac{5}{4}) + 7 b + 15$ .

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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.

$- 2 = 49 (- \frac{b}{4}) + 49 (- \frac{5}{4}) + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.1.2

Multiply $49 (- \frac{b}{4})$ .

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Step 2.3.4.2.1.1.2.1

Multiply $- 1$ by $49$ .

$- 2 = - 49 \frac{b}{4} + 49 (- \frac{5}{4}) + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.1.2.2

Combine $- 49$ and $\frac{b}{4}$ .

$- 2 = \frac{- 49 b}{4} + 49 (- \frac{5}{4}) + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = \frac{- 49 b}{4} + 49 (- \frac{5}{4}) + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.1.3

Multiply $49 (- \frac{5}{4})$ .

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Step 2.3.4.2.1.1.3.1

Multiply $- 1$ by $49$ .

$- 2 = \frac{- 49 b}{4} - 49 (\frac{5}{4}) + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.1.3.2

Combine $- 49$ and $\frac{5}{4}$ .

$- 2 = \frac{- 49 b}{4} + \frac{- 49 \cdot 5}{4} + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.1.3.3

Multiply $- 49$ by $5$ .

$- 2 = \frac{- 49 b}{4} + \frac{- 245}{4} + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = \frac{- 49 b}{4} + \frac{- 245}{4} + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.1.4

Simplify each term.

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Step 2.3.4.2.1.1.4.1

Move the negative in front of the fraction.

$- 2 = - \frac{49 b}{4} + \frac{- 245}{4} + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.1.4.2

Move the negative in front of the fraction.

$- 2 = - \frac{49 b}{4} - \frac{245}{4} + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = - \frac{49 b}{4} - \frac{245}{4} + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = - \frac{49 b}{4} - \frac{245}{4} + 7 b + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.2

To write $7 b$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$- 2 = - \frac{49 b}{4} + 7 b \cdot \frac{4}{4} - \frac{245}{4} + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.3

Combine $7 b$ and $\frac{4}{4}$ .

$- 2 = - \frac{49 b}{4} + \frac{7 b \cdot 4}{4} - \frac{245}{4} + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.4

Combine the numerators over the common denominator.

$- 2 = \frac{- 49 b + 7 b \cdot 4}{4} - \frac{245}{4} + 15$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.5

Combine the numerators over the common denominator.

$- 2 = 15 + \frac{- 49 b + 7 b \cdot 4 - 245}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.6

Multiply $4$ by $7$ .

$- 2 = 15 + \frac{- 49 b + 28 b - 245}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.7

Add $- 49 b$ and $28 b$ .

$- 2 = 15 + \frac{- 21 b - 245}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.8

Factor $7$ out of $- 21 b - 245$ .

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Step 2.3.4.2.1.8.1

Factor $7$ out of $- 21 b$ .

$- 2 = 15 + \frac{7 (- 3 b) - 245}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.8.2

Factor $7$ out of $- 245$ .

$- 2 = 15 + \frac{7 (- 3 b) + 7 (- 35)}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.8.3

Factor $7$ out of $7 (- 3 b) + 7 (- 35)$ .

$- 2 = 15 + \frac{7 (- 3 b - 35)}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = 15 + \frac{7 (- 3 b - 35)}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.9

To write $15$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$- 2 = 15 \cdot \frac{4}{4} + \frac{7 (- 3 b - 35)}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.10

Simplify terms.

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Step 2.3.4.2.1.10.1

Combine $15$ and $\frac{4}{4}$ .

$- 2 = \frac{15 \cdot 4}{4} + \frac{7 (- 3 b - 35)}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.10.2

Combine the numerators over the common denominator.

$- 2 = \frac{15 \cdot 4 + 7 (- 3 b - 35)}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = \frac{15 \cdot 4 + 7 (- 3 b - 35)}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.11

Simplify the numerator.

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Step 2.3.4.2.1.11.1

Multiply $15$ by $4$ .

$- 2 = \frac{60 + 7 (- 3 b - 35)}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.11.2

Apply the distributive property.

$- 2 = \frac{60 + 7 (- 3 b) + 7 \cdot - 35}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.11.3

Multiply $- 3$ by $7$ .

$- 2 = \frac{60 - 21 b + 7 \cdot - 35}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.11.4

Multiply $7$ by $- 35$ .

$- 2 = \frac{60 - 21 b - 245}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.11.5

Subtract $245$ from $60$ .

$- 2 = \frac{- 21 b - 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = \frac{- 21 b - 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.12

Simplify with factoring out.

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Step 2.3.4.2.1.12.1

Factor $- 1$ out of $- 21 b$ .

$- 2 = \frac{- (21 b) - 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.12.2

Rewrite $- 185$ as $- 1 (185)$ .

$- 2 = \frac{- (21 b) - 1 \cdot 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.12.3

Factor $- 1$ out of $- (21 b) - 1 (185)$ .

$- 2 = \frac{- (21 b + 185)}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.12.4

Simplify the expression.

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Step 2.3.4.2.1.12.4.1

Rewrite $- (21 b + 185)$ as $- 1 (21 b + 185)$ .

$- 2 = \frac{- 1 (21 b + 185)}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.2.1.12.4.2

Move the negative in front of the fraction.

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$10 = 25 a - 5 b + 15$

$c = 15$

Step 2.3.4.3

Replace all occurrences of $a$ in $10 = 25 a - 5 b + 15$ with $- \frac{b}{4} - \frac{5}{4}$ .

$10 = 25 (- \frac{b}{4} - \frac{5}{4}) - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4

Simplify the right side.

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Step 2.3.4.4.1

Simplify $25 (- \frac{b}{4} - \frac{5}{4}) - 5 b + 15$ .

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Step 2.3.4.4.1.1

Simplify each term.

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Step 2.3.4.4.1.1.1

Apply the distributive property.

$10 = 25 (- \frac{b}{4}) + 25 (- \frac{5}{4}) - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.1.2

Multiply $25 (- \frac{b}{4})$ .

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Step 2.3.4.4.1.1.2.1

Multiply $- 1$ by $25$ .

$10 = - 25 \frac{b}{4} + 25 (- \frac{5}{4}) - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.1.2.2

Combine $- 25$ and $\frac{b}{4}$ .

$10 = \frac{- 25 b}{4} + 25 (- \frac{5}{4}) - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = \frac{- 25 b}{4} + 25 (- \frac{5}{4}) - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.1.3

Multiply $25 (- \frac{5}{4})$ .

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Step 2.3.4.4.1.1.3.1

Multiply $- 1$ by $25$ .

$10 = \frac{- 25 b}{4} - 25 (\frac{5}{4}) - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.1.3.2

Combine $- 25$ and $\frac{5}{4}$ .

$10 = \frac{- 25 b}{4} + \frac{- 25 \cdot 5}{4} - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.1.3.3

Multiply $- 25$ by $5$ .

$10 = \frac{- 25 b}{4} + \frac{- 125}{4} - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = \frac{- 25 b}{4} + \frac{- 125}{4} - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.1.4

Simplify each term.

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Step 2.3.4.4.1.1.4.1

Move the negative in front of the fraction.

$10 = - \frac{25 b}{4} + \frac{- 125}{4} - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.1.4.2

Move the negative in front of the fraction.

$10 = - \frac{25 b}{4} - \frac{125}{4} - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = - \frac{25 b}{4} - \frac{125}{4} - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = - \frac{25 b}{4} - \frac{125}{4} - 5 b + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.2

To write $- 5 b$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$10 = - \frac{25 b}{4} - 5 b \cdot \frac{4}{4} - \frac{125}{4} + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.3

Combine $- 5 b$ and $\frac{4}{4}$ .

$10 = - \frac{25 b}{4} + \frac{- 5 b \cdot 4}{4} - \frac{125}{4} + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.4

Combine the numerators over the common denominator.

$10 = \frac{- 25 b - 5 b \cdot 4}{4} - \frac{125}{4} + 15$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.5

Combine the numerators over the common denominator.

$10 = 15 + \frac{- 25 b - 5 b \cdot 4 - 125}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.6

Multiply $4$ by $- 5$ .

$10 = 15 + \frac{- 25 b - 20 b - 125}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.7

Subtract $20 b$ from $- 25 b$ .

$10 = 15 + \frac{- 45 b - 125}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.8

Factor $5$ out of $- 45 b - 125$ .

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Step 2.3.4.4.1.8.1

Factor $5$ out of $- 45 b$ .

$10 = 15 + \frac{5 (- 9 b) - 125}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.8.2

Factor $5$ out of $- 125$ .

$10 = 15 + \frac{5 (- 9 b) + 5 (- 25)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.8.3

Factor $5$ out of $5 (- 9 b) + 5 (- 25)$ .

$10 = 15 + \frac{5 (- 9 b - 25)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = 15 + \frac{5 (- 9 b - 25)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.9

To write $15$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$10 = 15 \cdot \frac{4}{4} + \frac{5 (- 9 b - 25)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.10

Simplify terms.

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Step 2.3.4.4.1.10.1

Combine $15$ and $\frac{4}{4}$ .

$10 = \frac{15 \cdot 4}{4} + \frac{5 (- 9 b - 25)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.10.2

Combine the numerators over the common denominator.

$10 = \frac{15 \cdot 4 + 5 (- 9 b - 25)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = \frac{15 \cdot 4 + 5 (- 9 b - 25)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.11

Simplify the numerator.

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Step 2.3.4.4.1.11.1

Factor $5$ out of $15 \cdot 4 + 5 (- 9 b - 25)$ .

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Step 2.3.4.4.1.11.1.1

Factor $5$ out of $15 \cdot 4$ .

$10 = \frac{5 (3 \cdot 4) + 5 (- 9 b - 25)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.11.1.2

Factor $5$ out of $5 (3 \cdot 4) + 5 (- 9 b - 25)$ .

$10 = \frac{5 (3 \cdot 4 - 9 b - 25)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = \frac{5 (3 \cdot 4 - 9 b - 25)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.11.2

Multiply $3$ by $4$ .

$10 = \frac{5 (12 - 9 b - 25)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.11.3

Subtract $25$ from $12$ .

$10 = \frac{5 (- 9 b - 13)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = \frac{5 (- 9 b - 13)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.12

Simplify with factoring out.

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Step 2.3.4.4.1.12.1

Factor $- 1$ out of $- 9 b$ .

$10 = \frac{5 (- (9 b) - 13)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.12.2

Rewrite $- 13$ as $- 1 (13)$ .

$10 = \frac{5 (- (9 b) - 1 \cdot 13)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.12.3

Factor $- 1$ out of $- (9 b) - 1 (13)$ .

$10 = \frac{5 (- (9 b + 13))}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.12.4

Simplify the expression.

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Step 2.3.4.4.1.12.4.1

Rewrite $- (9 b + 13)$ as $- 1 (9 b + 13)$ .

$10 = \frac{5 (- 1 (9 b + 13))}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.4.4.1.12.4.2

Move the negative in front of the fraction.

$10 = - \frac{5 (9 b + 13)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = - \frac{5 (9 b + 13)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = - \frac{5 (9 b + 13)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = - \frac{5 (9 b + 13)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = - \frac{5 (9 b + 13)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$10 = - \frac{5 (9 b + 13)}{4}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5

Solve for $b$ in $10 = - \frac{5 (9 b + 13)}{4}$ .

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Step 2.3.5.1

Rewrite the equation as $- \frac{5 (9 b + 13)}{4} = 10$ .

$- \frac{5 (9 b + 13)}{4} = 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.2

Multiply both sides of the equation by $- \frac{4}{5}$ .

$- \frac{4}{5} \cdot (- \frac{5 (9 b + 13)}{4}) = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3

Simplify both sides of the equation.

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Step 2.3.5.3.1

Simplify the left side.

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Step 2.3.5.3.1.1

Simplify $- \frac{4}{5} (- \frac{5 (9 b + 13)}{4})$ .

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Step 2.3.5.3.1.1.1

Cancel the common factor of $4$ .

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Step 2.3.5.3.1.1.1.1

Move the leading negative in $- \frac{4}{5}$ into the numerator.

$\frac{- 4}{5} \cdot (- \frac{5 (9 b + 13)}{4}) = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.1.1.1.2

Move the leading negative in $- \frac{5 (9 b + 13)}{4}$ into the numerator.

$\frac{- 4}{5} \cdot \frac{- 5 (9 b + 13)}{4} = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.1.1.1.3

Factor $4$ out of $- 4$ .

$\frac{4 (- 1)}{5} \cdot \frac{- 5 (9 b + 13)}{4} = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.1.1.1.4

Cancel the common factor.

$\frac{4 \cdot - 1}{5} \cdot \frac{- 5 (9 b + 13)}{4} = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.1.1.1.5

Rewrite the expression.

$\frac{- 1}{5} \cdot (- 5 (9 b + 13)) = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$\frac{- 1}{5} \cdot (- 5 (9 b + 13)) = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.1.1.2

Cancel the common factor of $5$ .

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Step 2.3.5.3.1.1.2.1

Factor $5$ out of $- 5 (9 b + 13)$ .

$\frac{- 1}{5} \cdot (5 (- (9 b + 13))) = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.1.1.2.2

Cancel the common factor.

$\frac{- 1}{5} \cdot (5 (- (9 b + 13))) = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.1.1.2.3

Rewrite the expression.

$9 b + 13 = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$9 b + 13 = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.1.1.3

Multiply.

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Step 2.3.5.3.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 (9 b + 13) = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.1.1.3.2

Multiply $9 b + 13$ by $1$ .

$9 b + 13 = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$9 b + 13 = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$9 b + 13 = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$9 b + 13 = - \frac{4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.2

Simplify the right side.

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Step 2.3.5.3.2.1

Simplify $- \frac{4}{5} \cdot 10$ .

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Step 2.3.5.3.2.1.1

Cancel the common factor of $5$ .

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Step 2.3.5.3.2.1.1.1

Move the leading negative in $- \frac{4}{5}$ into the numerator.

$9 b + 13 = \frac{- 4}{5} \cdot 10$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.2.1.1.2

Factor $5$ out of $10$ .

$9 b + 13 = \frac{- 4}{5} \cdot (5 (2))$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.2.1.1.3

Cancel the common factor.

$9 b + 13 = \frac{- 4}{5} \cdot (5 \cdot 2)$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.2.1.1.4

Rewrite the expression.

$9 b + 13 = - 4 \cdot 2$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$9 b + 13 = - 4 \cdot 2$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.3.2.1.2

Multiply $- 4$ by $2$ .

$9 b + 13 = - 8$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$9 b + 13 = - 8$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$9 b + 13 = - 8$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$9 b + 13 = - 8$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.4

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

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Step 2.3.5.4.1

Subtract $13$ from both sides of the equation.

$9 b = - 8 - 13$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.4.2

Subtract $13$ from $- 8$ .

$9 b = - 21$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$9 b = - 21$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.5

Divide each term in $9 b = - 21$ by $9$ and simplify.

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Step 2.3.5.5.1

Divide each term in $9 b = - 21$ by $9$ .

$\frac{9 b}{9} = \frac{- 21}{9}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.5.2

Simplify the left side.

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Step 2.3.5.5.2.1

Cancel the common factor of $9$ .

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Step 2.3.5.5.2.1.1

Cancel the common factor.

$\frac{9 b}{9} = \frac{- 21}{9}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.5.2.1.2

Divide $b$ by $1$ .

$b = \frac{- 21}{9}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$b = \frac{- 21}{9}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$b = \frac{- 21}{9}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.5.3

Simplify the right side.

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Step 2.3.5.5.3.1

Cancel the common factor of $- 21$ and $9$ .

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Step 2.3.5.5.3.1.1

Factor $3$ out of $- 21$ .

$b = \frac{3 (- 7)}{9}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.5.3.1.2

Cancel the common factors.

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Step 2.3.5.5.3.1.2.1

Factor $3$ out of $9$ .

$b = \frac{3 \cdot - 7}{3 \cdot 3}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.5.3.1.2.2

Cancel the common factor.

$b = \frac{3 \cdot - 7}{3 \cdot 3}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.5.3.1.2.3

Rewrite the expression.

$b = \frac{- 7}{3}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$b = \frac{- 7}{3}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$b = \frac{- 7}{3}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.5.5.3.2

Move the negative in front of the fraction.

$b = - \frac{7}{3}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$b = - \frac{7}{3}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$b = - \frac{7}{3}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$b = - \frac{7}{3}$

$- 2 = - \frac{21 b + 185}{4}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.6

Replace all occurrences of $b$ with $- \frac{7}{3}$ in each equation.

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Step 2.3.6.1

Replace all occurrences of $b$ in $- 2 = - \frac{21 b + 185}{4}$ with $- \frac{7}{3}$ .

$- 2 = - \frac{21 (- \frac{7}{3}) + 185}{4}$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.6.2

Simplify the right side.

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Step 2.3.6.2.1

Simplify $- \frac{21 (- \frac{7}{3}) + 185}{4}$ .

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Step 2.3.6.2.1.1

Simplify the numerator.

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Step 2.3.6.2.1.1.1

Cancel the common factor of $3$ .

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Step 2.3.6.2.1.1.1.1

Move the leading negative in $- \frac{7}{3}$ into the numerator.

$- 2 = - \frac{21 (\frac{- 7}{3}) + 185}{4}$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.6.2.1.1.1.2

Factor $3$ out of $21$ .

$- 2 = - \frac{3 (7) (\frac{- 7}{3}) + 185}{4}$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.6.2.1.1.1.3

Cancel the common factor.

$- 2 = - \frac{3 \cdot (7 (\frac{- 7}{3})) + 185}{4}$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.6.2.1.1.1.4

Rewrite the expression.

$- 2 = - \frac{7 \cdot - 7 + 185}{4}$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$- 2 = - \frac{7 \cdot - 7 + 185}{4}$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.6.2.1.1.2

Multiply $7$ by $- 7$ .

$- 2 = - \frac{- 49 + 185}{4}$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.6.2.1.1.3

Add $- 49$ and $185$ .

$- 2 = - \frac{136}{4}$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$- 2 = - \frac{136}{4}$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.6.2.1.2

Simplify the expression.

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Step 2.3.6.2.1.2.1

Divide $136$ by $4$ .

$- 2 = - 1 \cdot 34$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.6.2.1.2.2

Multiply $- 1$ by $34$ .

$- 2 = - 34$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$- 2 = - 34$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$- 2 = - 34$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

$- 2 = - 34$

$b = - \frac{7}{3}$

$a = - \frac{b}{4} - \frac{5}{4}$

$c = 15$

Step 2.3.6.3

Replace all occurrences of $b$ in $a = - \frac{b}{4} - \frac{5}{4}$ with $- \frac{7}{3}$ .

$a = - \frac{- \frac{7}{3}}{4} - \frac{5}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4

Simplify the right side.

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Step 2.3.6.4.1

Simplify $- \frac{- \frac{7}{3}}{4} - \frac{5}{4}$ .

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Step 2.3.6.4.1.1

Combine the numerators over the common denominator.

$a = \frac{\frac{7}{3} - 5}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.2

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$a = \frac{\frac{7}{3} - 5 \cdot \frac{3}{3}}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.3

Combine $- 5$ and $\frac{3}{3}$ .

$a = \frac{\frac{7}{3} + \frac{- 5 \cdot 3}{3}}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.4

Combine the numerators over the common denominator.

$a = \frac{\frac{7 - 5 \cdot 3}{3}}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.5

Simplify the numerator.

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Step 2.3.6.4.1.5.1

Multiply $- 5$ by $3$ .

$a = \frac{\frac{7 - 15}{3}}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.5.2

Subtract $15$ from $7$ .

$a = \frac{\frac{- 8}{3}}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

$a = \frac{\frac{- 8}{3}}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.6

Move the negative in front of the fraction.

$a = \frac{- \frac{8}{3}}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.7

Multiply the numerator by the reciprocal of the denominator.

$a = - \frac{8}{3} \cdot \frac{1}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.8

Cancel the common factor of $4$ .

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Step 2.3.6.4.1.8.1

Move the leading negative in $- \frac{8}{3}$ into the numerator.

$a = \frac{- 8}{3} \cdot \frac{1}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.8.2

Factor $4$ out of $- 8$ .

$a = \frac{4 (- 2)}{3} \cdot \frac{1}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.8.3

Cancel the common factor.

$a = \frac{4 \cdot - 2}{3} \cdot \frac{1}{4}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.8.4

Rewrite the expression.

$a = \frac{- 2}{3}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

$a = \frac{- 2}{3}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.6.4.1.9

Move the negative in front of the fraction.

$a = - \frac{2}{3}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

$a = - \frac{2}{3}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

$a = - \frac{2}{3}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

$a = - \frac{2}{3}$

$- 2 = - 34$

$b = - \frac{7}{3}$

$c = 15$

Step 2.3.7

Since $- 2 = - 34$ 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 $x = - 5$ .

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Step 2.4.1.1

Simplify each term.

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Step 2.4.1.1.1

Raise $- 5$ to the power of $2$ .

$y = 0 \cdot 25 + (0) \cdot (- 5) + 0$

Step 2.4.1.1.2

Multiply $0$ by $25$ .

$y = 0 + (0) \cdot (- 5) + 0$

Step 2.4.1.1.3

Multiply $0$ by $- 5$ .

$y = 0 + 0 + 0$

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, $x = - 5$ . This check does not pass, since $y = 0$ and $y = 10$ . The function rule can't be quadratic.

$0 \neq 10$

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 $10 = a {(- 5)}^{3} + b {(- 5)}^{2} + c (- 5) + d$ .

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Step 3.3.1.1

Rewrite the equation as $a {(- 5)}^{3} + b {(- 5)}^{2} + c (- 5) + d = 10$ .

$a {(- 5)}^{3} + b {(- 5)}^{2} + c (- 5) + d = 10$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.1.2

Simplify each term.

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Step 3.3.1.2.1

Raise $- 5$ to the power of $3$ .

$a \cdot - 125 + b {(- 5)}^{2} + c (- 5) + d = 10$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.1.2.2

Move $- 125$ to the left of $a$ .

$- 125 \cdot a + b {(- 5)}^{2} + c (- 5) + d = 10$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.1.2.3

Raise $- 5$ to the power of $2$ .

$- 125 a + b \cdot 25 + c (- 5) + d = 10$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.1.2.4

Move $25$ to the left of $b$ .

$- 125 a + 25 \cdot b + c (- 5) + d = 10$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.1.2.5

Move $- 5$ to the left of $c$ .

$- 125 a + 25 b - 5 c + d = 10$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 125 a + 25 b - 5 c + d = 10$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

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 $125 a$ to both sides of the equation.

$25 b - 5 c + d = 10 + 125 a$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.1.3.2

Subtract $25 b$ from both sides of the equation.

$- 5 c + d = 10 + 125 a - 25 b$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.1.3.3

Add $5 c$ to both sides of the equation.

$d = 10 + 125 a - 25 b + 5 c$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$d = 10 + 125 a - 25 b + 5 c$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$d = 10 + 125 a - 25 b + 5 c$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2

Replace all occurrences of $d$ with $10 + 125 a - 25 b + 5 c$ in each equation.

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Step 3.3.2.1

Replace all occurrences of $d$ in $- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + d$ with $10 + 125 a - 25 b + 5 c$ .

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.2

Simplify $- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$ .

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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.

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.2.2

Simplify the right side.

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Step 3.3.2.2.2.1

Simplify $a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$ .

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Step 3.3.2.2.2.1.1

Combine the opposite terms in $a \cdot 7^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$ .

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Step 3.3.2.2.2.1.1.1

Reorder the factors in the terms $c (- 5)$ and $5 c$ .

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} - 5 c + 10 + 125 a - 25 b + 5 c$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.2.2.1.1.2

Add $- 5 c$ and $5 c$ .

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b + 0$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.2.2.1.1.3

Add $a \cdot 7^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b$ and $0$ .

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 2 = a \cdot 7^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

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 $7$ to the power of $3$ .

$- 2 = a \cdot 343 + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.2.2.1.2.2

Move $343$ to the left of $a$ .

$- 2 = 343 \cdot a + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.2.2.1.2.3

Raise $- 5$ to the power of $2$ .

$- 2 = 343 a + b \cdot 25 + 10 + 125 a - 25 b$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.2.2.1.2.4

Move $25$ to the left of $b$ .

$- 2 = 343 a + 25 b + 10 + 125 a - 25 b$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 2 = 343 a + 25 b + 10 + 125 a - 25 b$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

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 $343 a + 25 b + 10 + 125 a - 25 b$ .

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Step 3.3.2.2.2.1.3.1.1

Subtract $25 b$ from $25 b$ .

$- 2 = 343 a + 10 + 125 a + 0$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.2.2.1.3.1.2

Add $343 a + 10 + 125 a$ and $0$ .

$- 2 = 343 a + 10 + 125 a$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 2 = 343 a + 10 + 125 a$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.2.2.1.3.2

Add $343 a$ and $125 a$ .

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.3

Replace all occurrences of $d$ in $15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + d$ with $10 + 125 a - 25 b + 5 c$ .

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.4

Simplify $15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$ .

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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.

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.4.2

Simplify the right side.

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Step 3.3.2.4.2.1

Simplify $a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$ .

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Step 3.3.2.4.2.1.1

Combine the opposite terms in $a \cdot 0^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$ .

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Step 3.3.2.4.2.1.1.1

Reorder the factors in the terms $c (- 5)$ and $5 c$ .

$15 = a \cdot 0^{3} + b {(- 5)}^{2} - 5 c + 10 + 125 a - 25 b + 5 c$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.4.2.1.1.2

Add $- 5 c$ and $5 c$ .

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b + 0$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.4.2.1.1.3

Add $a \cdot 0^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b$ and $0$ .

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = a \cdot 0^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

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 $0$ .

$15 = a \cdot 0 + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.4.2.1.2.2

Multiply $a$ by $0$ .

$15 = 0 + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.4.2.1.2.3

Raise $- 5$ to the power of $2$ .

$15 = 0 + b \cdot 25 + 10 + 125 a - 25 b$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.4.2.1.2.4

Move $25$ to the left of $b$ .

$15 = 0 + 25 b + 10 + 125 a - 25 b$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = 0 + 25 b + 10 + 125 a - 25 b$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.4.2.1.3

Combine the opposite terms in $0 + 25 b + 10 + 125 a - 25 b$ .

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Step 3.3.2.4.2.1.3.1

Add $0$ and $25 b$ .

$15 = 25 b + 10 + 125 a - 25 b$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.4.2.1.3.2

Subtract $25 b$ from $25 b$ .

$15 = 10 + 125 a + 0$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.4.2.1.3.3

Add $10 + 125 a$ and $0$ .

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$

Step 3.3.2.5

Replace all occurrences of $d$ in $- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + d$ with $10 + 125 a - 25 b + 5 c$ .

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.2.6

Simplify $- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$ .

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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.

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.2.6.2

Simplify the right side.

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Step 3.3.2.6.2.1

Simplify $a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$ .

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Step 3.3.2.6.2.1.1

Combine the opposite terms in $a \cdot 4^{3} + b {(- 5)}^{2} + c (- 5) + 10 + 125 a - 25 b + 5 c$ .

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Step 3.3.2.6.2.1.1.1

Reorder the factors in the terms $c (- 5)$ and $5 c$ .

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} - 5 c + 10 + 125 a - 25 b + 5 c$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.2.6.2.1.1.2

Add $- 5 c$ and $5 c$ .

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b + 0$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.2.6.2.1.1.3

Add $a \cdot 4^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b$ and $0$ .

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = a \cdot 4^{3} + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

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 $4$ to the power of $3$ .

$- 5 = a \cdot 64 + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.2.6.2.1.2.2

Move $64$ to the left of $a$ .

$- 5 = 64 \cdot a + b {(- 5)}^{2} + 10 + 125 a - 25 b$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.2.6.2.1.2.3

Raise $- 5$ to the power of $2$ .

$- 5 = 64 a + b \cdot 25 + 10 + 125 a - 25 b$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.2.6.2.1.2.4

Move $25$ to the left of $b$ .

$- 5 = 64 a + 25 b + 10 + 125 a - 25 b$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = 64 a + 25 b + 10 + 125 a - 25 b$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

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 $64 a + 25 b + 10 + 125 a - 25 b$ .

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Step 3.3.2.6.2.1.3.1.1

Subtract $25 b$ from $25 b$ .

$- 5 = 64 a + 10 + 125 a + 0$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.2.6.2.1.3.1.2

Add $64 a + 10 + 125 a$ and $0$ .

$- 5 = 64 a + 10 + 125 a$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = 64 a + 10 + 125 a$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.2.6.2.1.3.2

Add $64 a$ and $125 a$ .

$- 5 = 189 a + 10$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = 189 a + 10$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = 189 a + 10$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = 189 a + 10$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = 189 a + 10$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$- 5 = 189 a + 10$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.3

Solve for $a$ in $- 5 = 189 a + 10$ .

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Step 3.3.3.1

Rewrite the equation as $189 a + 10 = - 5$ .

$189 a + 10 = - 5$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

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 $10$ from both sides of the equation.

$189 a = - 5 - 10$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.3.2.2

Subtract $10$ from $- 5$ .

$189 a = - 15$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$189 a = - 15$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.3.3

Divide each term in $189 a = - 15$ by $189$ and simplify.

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Step 3.3.3.3.1

Divide each term in $189 a = - 15$ by $189$ .

$\frac{189 a}{189} = \frac{- 15}{189}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

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 $189$ .

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Step 3.3.3.3.2.1.1

Cancel the common factor.

$\frac{189 a}{189} = \frac{- 15}{189}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{- 15}{189}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$a = \frac{- 15}{189}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$a = \frac{- 15}{189}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.3.3.3

Simplify the right side.

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Step 3.3.3.3.3.1

Cancel the common factor of $- 15$ and $189$ .

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Step 3.3.3.3.3.1.1

Factor $3$ out of $- 15$ .

$a = \frac{3 (- 5)}{189}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.3.3.3.1.2

Cancel the common factors.

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Step 3.3.3.3.3.1.2.1

Factor $3$ out of $189$ .

$a = \frac{3 \cdot - 5}{3 \cdot 63}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.3.3.3.1.2.2

Cancel the common factor.

$a = \frac{3 \cdot - 5}{3 \cdot 63}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.3.3.3.1.2.3

Rewrite the expression.

$a = \frac{- 5}{63}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$a = \frac{- 5}{63}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$a = \frac{- 5}{63}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.3.3.3.2

Move the negative in front of the fraction.

$a = - \frac{5}{63}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$a = - \frac{5}{63}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$a = - \frac{5}{63}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$a = - \frac{5}{63}$

$15 = 10 + 125 a$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4

Replace all occurrences of $a$ with $- \frac{5}{63}$ in each equation.

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Step 3.3.4.1

Replace all occurrences of $a$ in $15 = 10 + 125 a$ with $- \frac{5}{63}$ .

$15 = 10 + 125 (- \frac{5}{63})$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.2

Simplify the right side.

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Step 3.3.4.2.1

Simplify $10 + 125 (- \frac{5}{63})$ .

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Step 3.3.4.2.1.1

Simplify each term.

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Step 3.3.4.2.1.1.1

Multiply $125 (- \frac{5}{63})$ .

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Step 3.3.4.2.1.1.1.1

Multiply $- 1$ by $125$ .

$15 = 10 - 125 (\frac{5}{63})$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.2.1.1.1.2

Combine $- 125$ and $\frac{5}{63}$ .

$15 = 10 + \frac{- 125 \cdot 5}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.2.1.1.1.3

Multiply $- 125$ by $5$ .

$15 = 10 + \frac{- 625}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$15 = 10 + \frac{- 625}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.2.1.1.2

Move the negative in front of the fraction.

$15 = 10 - \frac{625}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$15 = 10 - \frac{625}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.2.1.2

To write $10$ as a fraction with a common denominator, multiply by $\frac{63}{63}$ .

$15 = 10 \cdot \frac{63}{63} - \frac{625}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.2.1.3

Combine $10$ and $\frac{63}{63}$ .

$15 = \frac{10 \cdot 63}{63} - \frac{625}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.2.1.4

Combine the numerators over the common denominator.

$15 = \frac{10 \cdot 63 - 625}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.2.1.5

Simplify the numerator.

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Step 3.3.4.2.1.5.1

Multiply $10$ by $63$ .

$15 = \frac{630 - 625}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.2.1.5.2

Subtract $625$ from $630$ .

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$- 2 = 468 a + 10$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.3

Replace all occurrences of $a$ in $- 2 = 468 a + 10$ with $- \frac{5}{63}$ .

$- 2 = 468 (- \frac{5}{63}) + 10$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4

Simplify the right side.

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Step 3.3.4.4.1

Simplify $468 (- \frac{5}{63}) + 10$ .

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Step 3.3.4.4.1.1

Simplify each term.

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Step 3.3.4.4.1.1.1

Cancel the common factor of $9$ .

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Step 3.3.4.4.1.1.1.1

Move the leading negative in $- \frac{5}{63}$ into the numerator.

$- 2 = 468 (\frac{- 5}{63}) + 10$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.1.1.2

Factor $9$ out of $468$ .

$- 2 = 9 (52) (\frac{- 5}{63}) + 10$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.1.1.3

Factor $9$ out of $63$ .

$- 2 = 9 \cdot (52 (\frac{- 5}{9 \cdot 7})) + 10$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.1.1.4

Cancel the common factor.

$- 2 = 9 \cdot (52 (\frac{- 5}{9 \cdot 7})) + 10$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.1.1.5

Rewrite the expression.

$- 2 = 52 (\frac{- 5}{7}) + 10$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

$- 2 = 52 (\frac{- 5}{7}) + 10$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.1.2

Combine $52$ and $\frac{- 5}{7}$ .

$- 2 = \frac{52 \cdot - 5}{7} + 10$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.1.3

Multiply $52$ by $- 5$ .

$- 2 = \frac{- 260}{7} + 10$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.1.4

Move the negative in front of the fraction.

$- 2 = - \frac{260}{7} + 10$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

$- 2 = - \frac{260}{7} + 10$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.2

To write $10$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$- 2 = - \frac{260}{7} + 10 \cdot \frac{7}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.3

Combine $10$ and $\frac{7}{7}$ .

$- 2 = - \frac{260}{7} + \frac{10 \cdot 7}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.4

Combine the numerators over the common denominator.

$- 2 = \frac{- 260 + 10 \cdot 7}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.5

Simplify the numerator.

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Step 3.3.4.4.1.5.1

Multiply $10$ by $7$ .

$- 2 = \frac{- 260 + 70}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.5.2

Add $- 260$ and $70$ .

$- 2 = \frac{- 190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

$- 2 = \frac{- 190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.4.1.6

Move the negative in front of the fraction.

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + 125 a - 25 b + 5 c$

Step 3.3.4.5

Replace all occurrences of $a$ in $d = 10 + 125 a - 25 b + 5 c$ with $- \frac{5}{63}$ .

$d = 10 + 125 (- \frac{5}{63}) - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

Step 3.3.4.6

Simplify the right side.

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Step 3.3.4.6.1

Simplify $10 + 125 (- \frac{5}{63}) - 25 b + 5 c$ .

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Step 3.3.4.6.1.1

Simplify each term.

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Step 3.3.4.6.1.1.1

Multiply $125 (- \frac{5}{63})$ .

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Step 3.3.4.6.1.1.1.1

Multiply $- 1$ by $125$ .

$d = 10 - 125 (\frac{5}{63}) - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

Step 3.3.4.6.1.1.1.2

Combine $- 125$ and $\frac{5}{63}$ .

$d = 10 + \frac{- 125 \cdot 5}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

Step 3.3.4.6.1.1.1.3

Multiply $- 125$ by $5$ .

$d = 10 + \frac{- 625}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 + \frac{- 625}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

Step 3.3.4.6.1.1.2

Move the negative in front of the fraction.

$d = 10 - \frac{625}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = 10 - \frac{625}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

Step 3.3.4.6.1.2

To write $10$ as a fraction with a common denominator, multiply by $\frac{63}{63}$ .

$d = 10 \cdot \frac{63}{63} - \frac{625}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

Step 3.3.4.6.1.3

Combine $10$ and $\frac{63}{63}$ .

$d = \frac{10 \cdot 63}{63} - \frac{625}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

Step 3.3.4.6.1.4

Combine the numerators over the common denominator.

$d = \frac{10 \cdot 63 - 625}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

Step 3.3.4.6.1.5

Simplify the numerator.

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Step 3.3.4.6.1.5.1

Multiply $10$ by $63$ .

$d = \frac{630 - 625}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

Step 3.3.4.6.1.5.2

Subtract $625$ from $630$ .

$d = \frac{5}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = \frac{5}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = \frac{5}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = \frac{5}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

$d = \frac{5}{63} - 25 b + 5 c$

$- 2 = - \frac{190}{7}$

$15 = \frac{5}{63}$

$a = - \frac{5}{63}$

Step 3.3.5

Since $- 2 = - \frac{190}{7}$ 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 $x = - 5$ .

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Step 3.4.1.1

Simplify each term.

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Step 3.4.1.1.1

Raise $- 5$ to the power of $3$ .

$y = 0 \cdot - 125 + (0) \cdot (- 5^{2}) + (0) \cdot ((- 5) \cdot 0)$

Step 3.4.1.1.2

Multiply $0$ by $- 125$ .

$y = 0 + (0) \cdot (- 5^{2}) + (0) \cdot ((- 5) \cdot 0)$

Step 3.4.1.1.3

Raise $5$ to the power of $2$ .

$y = 0 + 0 \cdot (- 1 \cdot 25) + (0) \cdot ((- 5) \cdot 0)$

Step 3.4.1.1.4

Multiply $0 (- 1 \cdot 25)$ .

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Step 3.4.1.1.4.1

Multiply $- 1$ by $25$ .

$y = 0 + 0 \cdot - 25 + (0) \cdot ((- 5) \cdot 0)$

Step 3.4.1.1.4.2

Multiply $0$ by $- 25$ .

$y = 0 + 0 + (0) \cdot ((- 5) \cdot 0)$

Step 3.4.1.1.5

Multiply $(0) ((- 5) \cdot 0)$ .

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Step 3.4.1.1.5.1

Multiply $- 5$ 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, $x = - 5$ . This check does not pass, since $y = 0$ and $y = 10$ . The function rule can't be cubic.

$0 \neq 10$

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.