Pre-Algebra Examples

$\begin{array}{c} x & y \\ 1 & 5 \\ 2 & 20 \\ 3 & 45 \end{array}$

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 $y = a x + b$ .

$5 = a (1) + b 20 = a (2) + b 45 = a (3) + b$

Step 1.3

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

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Step 1.3.1

Solve for $a$ in $5 = a + b$ .

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Step 1.3.1.1

Rewrite the equation as $a + b = 5$ .

$a + b = 5$

$20 = a (2) + b$

$45 = a (3) + b$

Step 1.3.1.2

Subtract $b$ from both sides of the equation.

$a = 5 - b$

$20 = a (2) + b$

$45 = a (3) + b$

$a = 5 - b$

$20 = a (2) + b$

$45 = a (3) + b$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $a$ in $20 = a (2) + b$ with $5 - b$ .

$20 = (5 - b) (2) + b$

$a = 5 - b$

$45 = a (3) + b$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $(5 - b) (2) + b$ .

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Step 1.3.2.2.1.1

Simplify each term.

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Step 1.3.2.2.1.1.1

Apply the distributive property.

$20 = 5 \cdot 2 - b \cdot 2 + b$

$a = 5 - b$

$45 = a (3) + b$

Step 1.3.2.2.1.1.2

Multiply $5$ by $2$ .

$20 = 10 - b \cdot 2 + b$

$a = 5 - b$

$45 = a (3) + b$

Step 1.3.2.2.1.1.3

Multiply $2$ by $- 1$ .

$20 = 10 - 2 b + b$

$a = 5 - b$

$45 = a (3) + b$

$20 = 10 - 2 b + b$

$a = 5 - b$

$45 = a (3) + b$

Step 1.3.2.2.1.2

Add $- 2 b$ and $b$ .

$20 = 10 - b$

$a = 5 - b$

$45 = a (3) + b$

$20 = 10 - b$

$a = 5 - b$

$45 = a (3) + b$

$20 = 10 - b$

$a = 5 - b$

$45 = a (3) + b$

Step 1.3.2.3

Replace all occurrences of $a$ in $45 = a (3) + b$ with $5 - b$ .

$45 = (5 - b) (3) + b$

$20 = 10 - b$

$a = 5 - b$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Simplify $(5 - b) (3) + b$ .

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Step 1.3.2.4.1.1

Simplify each term.

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Step 1.3.2.4.1.1.1

Apply the distributive property.

$45 = 5 \cdot 3 - b \cdot 3 + b$

$20 = 10 - b$

$a = 5 - b$

Step 1.3.2.4.1.1.2

Multiply $5$ by $3$ .

$45 = 15 - b \cdot 3 + b$

$20 = 10 - b$

$a = 5 - b$

Step 1.3.2.4.1.1.3

Multiply $3$ by $- 1$ .

$45 = 15 - 3 b + b$

$20 = 10 - b$

$a = 5 - b$

$45 = 15 - 3 b + b$

$20 = 10 - b$

$a = 5 - b$

Step 1.3.2.4.1.2

Add $- 3 b$ and $b$ .

$45 = 15 - 2 b$

$20 = 10 - b$

$a = 5 - b$

$45 = 15 - 2 b$

$20 = 10 - b$

$a = 5 - b$

$45 = 15 - 2 b$

$20 = 10 - b$

$a = 5 - b$

$45 = 15 - 2 b$

$20 = 10 - b$

$a = 5 - b$

Step 1.3.3

Solve for $b$ in $45 = 15 - 2 b$ .

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Step 1.3.3.1

Rewrite the equation as $15 - 2 b = 45$ .

$15 - 2 b = 45$

$20 = 10 - b$

$a = 5 - b$

Step 1.3.3.2

Move all terms not containing $b$ 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.

$- 2 b = 45 - 15$

$20 = 10 - b$

$a = 5 - b$

Step 1.3.3.2.2

Subtract $15$ from $45$ .

$- 2 b = 30$

$20 = 10 - b$

$a = 5 - b$

$- 2 b = 30$

$20 = 10 - b$

$a = 5 - b$

Step 1.3.3.3

Divide each term in $- 2 b = 30$ by $- 2$ and simplify.

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Step 1.3.3.3.1

Divide each term in $- 2 b = 30$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{30}{- 2}$

$20 = 10 - b$

$a = 5 - b$

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.

$\frac{- 2 b}{- 2} = \frac{30}{- 2}$

$20 = 10 - b$

$a = 5 - b$

Step 1.3.3.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{30}{- 2}$

$20 = 10 - b$

$a = 5 - b$

$b = \frac{30}{- 2}$

$20 = 10 - b$

$a = 5 - b$

$b = \frac{30}{- 2}$

$20 = 10 - b$

$a = 5 - b$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide $30$ by $- 2$ .

$b = - 15$

$20 = 10 - b$

$a = 5 - b$

$b = - 15$

$20 = 10 - b$

$a = 5 - b$

$b = - 15$

$20 = 10 - b$

$a = 5 - b$

$b = - 15$

$20 = 10 - b$

$a = 5 - b$

Step 1.3.4

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

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Step 1.3.4.1

Replace all occurrences of $b$ in $20 = 10 - b$ with $- 15$ .

$20 = 10 - (- 15)$

$b = - 15$

$a = 5 - b$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $10 - (- 15)$ .

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Step 1.3.4.2.1.1

Multiply $- 1$ by $- 15$ .

$20 = 10 + 15$

$b = - 15$

$a = 5 - b$

Step 1.3.4.2.1.2

Add $10$ and $15$ .

$20 = 25$

$b = - 15$

$a = 5 - b$

$20 = 25$

$b = - 15$

$a = 5 - b$

$20 = 25$

$b = - 15$

$a = 5 - b$

Step 1.3.4.3

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

$a = 5 - (- 15)$

$20 = 25$

$b = - 15$

Step 1.3.4.4

Simplify the right side.

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Step 1.3.4.4.1

Simplify $5 - (- 15)$ .

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Step 1.3.4.4.1.1

Multiply $- 1$ by $- 15$ .

$a = 5 + 15$

$20 = 25$

$b = - 15$

Step 1.3.4.4.1.2

Add $5$ and $15$ .

$a = 20$

$20 = 25$

$b = - 15$

$a = 20$

$20 = 25$

$b = - 15$

$a = 20$

$20 = 25$

$b = - 15$

$a = 20$

$20 = 25$

$b = - 15$

Step 1.3.5

Since $20 = 25$ 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 $a$ in $5 = a + b + c$ .

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Step 2.3.1.1

Rewrite the equation as $a + b + c = 5$ .

$a + b + c = 5$

$20 = a \cdot 2^{2} + b (2) + c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.1.2

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

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Step 2.3.1.2.1

Subtract $b$ from both sides of the equation.

$a + c = 5 - b$

$20 = a \cdot 2^{2} + b (2) + c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.1.2.2

Subtract $c$ from both sides of the equation.

$a = 5 - b - c$

$20 = a \cdot 2^{2} + b (2) + c$

$45 = a \cdot 3^{2} + b (3) + c$

$a = 5 - b - c$

$20 = a \cdot 2^{2} + b (2) + c$

$45 = a \cdot 3^{2} + b (3) + c$

$a = 5 - b - c$

$20 = a \cdot 2^{2} + b (2) + c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.2

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

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Step 2.3.2.1

Replace all occurrences of $a$ in $20 = a \cdot 2^{2} + b (2) + c$ with $5 - b - c$ .

$20 = (5 - b - c) \cdot 2^{2} + b (2) + c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.2.2

Simplify the right side.

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Step 2.3.2.2.1

Simplify $(5 - b - c) \cdot 2^{2} + b (2) + c$ .

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Step 2.3.2.2.1.1

Simplify each term.

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Step 2.3.2.2.1.1.1

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

$20 = (5 - b - c) \cdot 4 + b (2) + c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.2.2.1.1.2

Apply the distributive property.

$20 = 5 \cdot 4 - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.2.2.1.1.3

Simplify.

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Step 2.3.2.2.1.1.3.1

Multiply $5$ by $4$ .

$20 = 20 - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.2.2.1.1.3.2

Multiply $4$ by $- 1$ .

$20 = 20 - 4 b - c \cdot 4 + b (2) + c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.2.2.1.1.3.3

Multiply $4$ by $- 1$ .

$20 = 20 - 4 b - 4 c + b (2) + c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

$20 = 20 - 4 b - 4 c + b (2) + c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.2.2.1.1.4

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

$20 = 20 - 4 b - 4 c + 2 b + c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

$20 = 20 - 4 b - 4 c + 2 b + c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.2.2.1.2

Simplify by adding terms.

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Step 2.3.2.2.1.2.1

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

$20 = 20 - 2 b - 4 c + c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.2.2.1.2.2

Add $- 4 c$ and $c$ .

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$45 = a \cdot 3^{2} + b (3) + c$

Step 2.3.2.3

Replace all occurrences of $a$ in $45 = a \cdot 3^{2} + b (3) + c$ with $5 - b - c$ .

$45 = (5 - b - c) \cdot 3^{2} + b (3) + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.2.4

Simplify the right side.

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Step 2.3.2.4.1

Simplify $(5 - b - c) \cdot 3^{2} + b (3) + c$ .

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Step 2.3.2.4.1.1

Simplify each term.

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Step 2.3.2.4.1.1.1

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

$45 = (5 - b - c) \cdot 9 + b (3) + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.2.4.1.1.2

Apply the distributive property.

$45 = 5 \cdot 9 - b \cdot 9 - c \cdot 9 + b (3) + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.2.4.1.1.3

Simplify.

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Step 2.3.2.4.1.1.3.1

Multiply $5$ by $9$ .

$45 = 45 - b \cdot 9 - c \cdot 9 + b (3) + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.2.4.1.1.3.2

Multiply $9$ by $- 1$ .

$45 = 45 - 9 b - c \cdot 9 + b (3) + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.2.4.1.1.3.3

Multiply $9$ by $- 1$ .

$45 = 45 - 9 b - 9 c + b (3) + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$45 = 45 - 9 b - 9 c + b (3) + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.2.4.1.1.4

Move $3$ to the left of $b$ .

$45 = 45 - 9 b - 9 c + 3 b + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$45 = 45 - 9 b - 9 c + 3 b + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.2.4.1.2

Simplify by adding terms.

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Step 2.3.2.4.1.2.1

Add $- 9 b$ and $3 b$ .

$45 = 45 - 6 b - 9 c + c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.2.4.1.2.2

Add $- 9 c$ and $c$ .

$45 = 45 - 6 b - 8 c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$45 = 45 - 6 b - 8 c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$45 = 45 - 6 b - 8 c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$45 = 45 - 6 b - 8 c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$45 = 45 - 6 b - 8 c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3

Solve for $b$ in $45 = 45 - 6 b - 8 c$ .

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Step 2.3.3.1

Rewrite the equation as $45 - 6 b - 8 c = 45$ .

$45 - 6 b - 8 c = 45$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3.2

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

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Step 2.3.3.2.1

Subtract $45$ from both sides of the equation.

$- 6 b - 8 c = 45 - 45$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3.2.2

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

$- 6 b = 45 - 45 + 8 c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3.2.3

Subtract $45$ from $45$ .

$- 6 b = 0 + 8 c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3.2.4

Add $0$ and $8 c$ .

$- 6 b = 8 c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$- 6 b = 8 c$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3.3

Divide each term in $- 6 b = 8 c$ by $- 6$ and simplify.

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Step 2.3.3.3.1

Divide each term in $- 6 b = 8 c$ by $- 6$ .

$\frac{- 6 b}{- 6} = \frac{8 c}{- 6}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

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 $- 6$ .

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Step 2.3.3.3.2.1.1

Cancel the common factor.

$\frac{- 6 b}{- 6} = \frac{8 c}{- 6}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{8 c}{- 6}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{8 c}{- 6}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{8 c}{- 6}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3.3.3

Simplify the right side.

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Step 2.3.3.3.3.1

Cancel the common factor of $8$ and $- 6$ .

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Step 2.3.3.3.3.1.1

Factor $2$ out of $8 c$ .

$b = \frac{2 (4 c)}{- 6}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3.3.3.1.2

Cancel the common factors.

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Step 2.3.3.3.3.1.2.1

Factor $2$ out of $- 6$ .

$b = \frac{2 (4 c)}{2 (- 3)}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3.3.3.1.2.2

Cancel the common factor.

$b = \frac{2 (4 c)}{2 \cdot - 3}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3.3.3.1.2.3

Rewrite the expression.

$b = \frac{4 c}{- 3}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{4 c}{- 3}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = \frac{4 c}{- 3}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.3.3.3.2

Move the negative in front of the fraction.

$b = - \frac{4 c}{3}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = - \frac{4 c}{3}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = - \frac{4 c}{3}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

$b = - \frac{4 c}{3}$

$20 = 20 - 2 b - 3 c$

$a = 5 - b - c$

Step 2.3.4

Replace all occurrences of $b$ with $- \frac{4 c}{3}$ in each equation.

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Step 2.3.4.1

Replace all occurrences of $b$ in $20 = 20 - 2 b - 3 c$ with $- \frac{4 c}{3}$ .

$20 = 20 - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2

Simplify the right side.

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Step 2.3.4.2.1

Simplify $20 - 2 (- \frac{4 c}{3}) - 3 c$ .

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Step 2.3.4.2.1.1

Multiply $- 2 (- \frac{4 c}{3})$ .

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Step 2.3.4.2.1.1.1

Multiply $- 1$ by $- 2$ .

$20 = 20 + 2 (\frac{4 c}{3}) - 3 c$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.1.2

Combine $2$ and $\frac{4 c}{3}$ .

$20 = 20 + \frac{2 (4 c)}{3} - 3 c$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.1.3

Multiply $4$ by $2$ .

$20 = 20 + \frac{8 c}{3} - 3 c$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

$20 = 20 + \frac{8 c}{3} - 3 c$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.2

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

$20 = 20 + \frac{8 c}{3} - 3 c \cdot \frac{3}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.3

Simplify terms.

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Step 2.3.4.2.1.3.1

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

$20 = 20 + \frac{8 c}{3} + \frac{- 3 c \cdot 3}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.3.2

Combine the numerators over the common denominator.

$20 = 20 + \frac{8 c - 3 c \cdot 3}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

$20 = 20 + \frac{8 c - 3 c \cdot 3}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.4

Simplify each term.

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Step 2.3.4.2.1.4.1

Simplify the numerator.

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Step 2.3.4.2.1.4.1.1

Factor $c$ out of $8 c - 3 c \cdot 3$ .

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Step 2.3.4.2.1.4.1.1.1

Factor $c$ out of $8 c$ .

$20 = 20 + \frac{c \cdot 8 - 3 c \cdot 3}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.4.1.1.2

Factor $c$ out of $- 3 c \cdot 3$ .

$20 = 20 + \frac{c \cdot 8 + c (- 3 \cdot 3)}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.4.1.1.3

Factor $c$ out of $c \cdot 8 + c (- 3 \cdot 3)$ .

$20 = 20 + \frac{c (8 - 3 \cdot 3)}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

$20 = 20 + \frac{c (8 - 3 \cdot 3)}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.4.1.2

Multiply $- 3$ by $3$ .

$20 = 20 + \frac{c (8 - 9)}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.4.1.3

Subtract $9$ from $8$ .

$20 = 20 + \frac{c \cdot - 1}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

$20 = 20 + \frac{c \cdot - 1}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.4.2

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

$20 = 20 + \frac{- 1 \cdot c}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.2.1.4.3

Move the negative in front of the fraction.

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 5 - b - c$

Step 2.3.4.3

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

$a = 5 - (- \frac{4 c}{3}) - c$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.4.4

Simplify the right side.

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Step 2.3.4.4.1

Simplify $5 - (- \frac{4 c}{3}) - c$ .

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Step 2.3.4.4.1.1

Multiply $- (- \frac{4 c}{3})$ .

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Step 2.3.4.4.1.1.1

Multiply $- 1$ by $- 1$ .

$a = 5 + 1 (\frac{4 c}{3}) - c$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.4.4.1.1.2

Multiply $\frac{4 c}{3}$ by $1$ .

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

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

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

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.4.4.1.2

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

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

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.4.4.1.3

Simplify terms.

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Step 2.3.4.4.1.3.1

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

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

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.4.4.1.3.2

Combine the numerators over the common denominator.

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

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

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

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.4.4.1.4

Simplify each term.

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Step 2.3.4.4.1.4.1

Simplify the numerator.

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Step 2.3.4.4.1.4.1.1

Factor $c$ out of $4 c - c \cdot 3$ .

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Step 2.3.4.4.1.4.1.1.1

Factor $c$ out of $4 c$ .

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

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.4.4.1.4.1.1.2

Factor $c$ out of $- c \cdot 3$ .

$a = 5 + \frac{c \cdot 4 + c (- 1 \cdot 3)}{3}$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.4.4.1.4.1.1.3

Factor $c$ out of $c \cdot 4 + c (- 1 \cdot 3)$ .

$a = 5 + \frac{c (4 - 1 \cdot 3)}{3}$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 5 + \frac{c (4 - 1 \cdot 3)}{3}$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.4.4.1.4.1.2

Multiply $- 1$ by $3$ .

$a = 5 + \frac{c (4 - 3)}{3}$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.4.4.1.4.1.3

Subtract $3$ from $4$ .

$a = 5 + \frac{c \cdot 1}{3}$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 5 + \frac{c \cdot 1}{3}$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.4.4.1.4.2

Multiply $c$ by $1$ .

$a = 5 + \frac{c}{3}$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 5 + \frac{c}{3}$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 5 + \frac{c}{3}$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 5 + \frac{c}{3}$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

$a = 5 + \frac{c}{3}$

$20 = 20 - \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.5

Solve for $c$ in $20 = 20 - \frac{c}{3}$ .

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Step 2.3.5.1

Rewrite the equation as $20 - \frac{c}{3} = 20$ .

$20 - \frac{c}{3} = 20$

$a = 5 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.5.2

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

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Step 2.3.5.2.1

Subtract $20$ from both sides of the equation.

$- \frac{c}{3} = 20 - 20$

$a = 5 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.5.2.2

Subtract $20$ from $20$ .

$- \frac{c}{3} = 0$

$a = 5 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

$- \frac{c}{3} = 0$

$a = 5 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.5.3

Set the numerator equal to zero.

$c = 0$

$a = 5 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

$c = 0$

$a = 5 + \frac{c}{3}$

$b = - \frac{4 c}{3}$

Step 2.3.6

Replace all occurrences of $c$ with $0$ in each equation.

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Step 2.3.6.1

Replace all occurrences of $c$ in $a = 5 + \frac{c}{3}$ with $0$ .

$a = 5 + \frac{0}{3}$

$c = 0$

$b = - \frac{4 c}{3}$

Step 2.3.6.2

Simplify the right side.

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Step 2.3.6.2.1

Simplify $5 + \frac{0}{3}$ .

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Step 2.3.6.2.1.1

Divide $0$ by $3$ .

$a = 5 + 0$

$c = 0$

$b = - \frac{4 c}{3}$

Step 2.3.6.2.1.2

Add $5$ and $0$ .

$a = 5$

$c = 0$

$b = - \frac{4 c}{3}$

$a = 5$

$c = 0$

$b = - \frac{4 c}{3}$

$a = 5$

$c = 0$

$b = - \frac{4 c}{3}$

Step 2.3.6.3

Replace all occurrences of $c$ in $b = - \frac{4 c}{3}$ with $0$ .

$b = - \frac{4 (0)}{3}$

$a = 5$

$c = 0$

Step 2.3.6.4

Simplify the right side.

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Step 2.3.6.4.1

Simplify $- \frac{4 (0)}{3}$ .

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Step 2.3.6.4.1.1

Cancel the common factor of $0$ and $3$ .

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Step 2.3.6.4.1.1.1

Factor $3$ out of $4 (0)$ .

$b = - \frac{3 (4 \cdot (0))}{3}$

$a = 5$

$c = 0$

Step 2.3.6.4.1.1.2

Cancel the common factors.

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Step 2.3.6.4.1.1.2.1

Factor $3$ out of $3$ .

$b = - \frac{3 (4 \cdot (0))}{3 (1)}$

$a = 5$

$c = 0$

Step 2.3.6.4.1.1.2.2

Cancel the common factor.

$b = - \frac{3 (4 \cdot (0))}{3 \cdot 1}$

$a = 5$

$c = 0$

Step 2.3.6.4.1.1.2.3

Rewrite the expression.

$b = - \frac{4 \cdot (0)}{1}$

$a = 5$

$c = 0$

Step 2.3.6.4.1.1.2.4

Divide $4 \cdot (0)$ by $1$ .

$b = - (4 \cdot (0))$

$a = 5$

$c = 0$

$b = - (4 \cdot (0))$

$a = 5$

$c = 0$

$b = - (4 \cdot (0))$

$a = 5$

$c = 0$

Step 2.3.6.4.1.2

Multiply $- (4 \cdot (0))$ .

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Step 2.3.6.4.1.2.1

Multiply $4$ by $0$ .

$b = - 0$

$a = 5$

$c = 0$

Step 2.3.6.4.1.2.2

Multiply $- 1$ by $0$ .

$b = 0$

$a = 5$

$c = 0$

$b = 0$

$a = 5$

$c = 0$

$b = 0$

$a = 5$

$c = 0$

$b = 0$

$a = 5$

$c = 0$

$b = 0$

$a = 5$

$c = 0$

Step 2.3.7

List all of the solutions.

$b = 0, a = 5, c = 0$

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 $a = 5$ , $b = 0$ , $c = 0$ , and $x = 1$ .

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Step 2.4.1.1

Simplify each term.

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Step 2.4.1.1.1

One to any power is one.

$y = 5 \cdot 1 + (0) \cdot (1) + 0$

Step 2.4.1.1.2

Multiply $5$ by $1$ .

$y = 5 + (0) \cdot (1) + 0$

Step 2.4.1.1.3

Multiply $0$ by $1$ .

$y = 5 + 0 + 0$

Step 2.4.1.2

Simplify by adding numbers.

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Step 2.4.1.2.1

Add $5$ and $0$ .

$y = 5 + 0$

Step 2.4.1.2.2

Add $5$ and $0$ .

$y = 5$

Step 2.4.2

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 5$ and $y = 5$ .

$5 = 5$

Step 2.4.3

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 5$ , $b = 0$ , $c = 0$ , and $x = 2$ .

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Step 2.4.3.1

Simplify each term.

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Step 2.4.3.1.1

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

$y = 5 \cdot 4 + (0) \cdot (2) + 0$

Step 2.4.3.1.2

Multiply $5$ by $4$ .

$y = 20 + (0) \cdot (2) + 0$

Step 2.4.3.1.3

Multiply $0$ by $2$ .

$y = 20 + 0 + 0$

Step 2.4.3.2

Simplify by adding numbers.

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Step 2.4.3.2.1

Add $20$ and $0$ .

$y = 20 + 0$

Step 2.4.3.2.2

Add $20$ and $0$ .

$y = 20$

Step 2.4.4

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 20$ and $y = 20$ .

$20 = 20$

Step 2.4.5

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 5$ , $b = 0$ , $c = 0$ , and $x = 3$ .

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Step 2.4.5.1

Simplify each term.

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Step 2.4.5.1.1

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

$y = 5 \cdot 9 + (0) \cdot (3) + 0$

Step 2.4.5.1.2

Multiply $5$ by $9$ .

$y = 45 + (0) \cdot (3) + 0$

Step 2.4.5.1.3

Multiply $0$ by $3$ .

$y = 45 + 0 + 0$

Step 2.4.5.2

Simplify by adding numbers.

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Step 2.4.5.2.1

Add $45$ and $0$ .

$y = 45 + 0$

Step 2.4.5.2.2

Add $45$ and $0$ .

$y = 45$

Step 2.4.6

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 45$ and $y = 45$ .

$45 = 45$

Step 2.4.7

Since $y = y$ for the corresponding $x$ values, the function is quadratic.

The function is quadratic

Step 3

Since all $y = y$ , the function is quadratic and follows the form $y = 5 x^{2}$ .

$y = 5 x^{2}$