Find the Function Rule table[[x,y],[1,5],[5,10],[9,15]]

$\begin{array}{c} x & y \\ 1 & 5 \\ 5 & 10 \\ 9 & 15 \end{array}$

Check if the function rule is linear.

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

$y = a x + b$

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

$5 = a (1) + b 10 = a (5) + b 15 = a (9) + b$

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

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Simplify each equation.

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Multiply $a$ by $1$ .

$5 = a + b$

$10 = a (5) + b$

$15 = a (9) + b$

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

$5 = a + b$

$10 = 5 a + b$

$15 = a (9) + b$

Move $9$ to the left of $a$ .

$5 = a + b$

$10 = 5 a + b$

$15 = 9 a + b$

$5 = a + b$

$10 = 5 a + b$

$15 = 9 a + b$

Solve for $a$ in the first equation.

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

$a + b = 5$

$10 = 5 a + b$

$15 = 9 a + b$

Subtract $b$ from both sides of the equation.

$a = 5 - b$

$10 = 5 a + b$

$15 = 9 a + b$

$a = 5 - b$

$10 = 5 a + b$

$15 = 9 a + b$

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

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

$a = 5 - b$

$10 = 5 (5 - b) + b$

$15 = 9 a + b$

Replace all occurrences of $a$ in $15 = 9 a + b$ with $5 - b$ .

$a = 5 - b$

$10 = 5 (5 - b) + b$

$15 = 9 (5 - b) + b$

$a = 5 - b$

$10 = 5 (5 - b) + b$

$15 = 9 (5 - b) + b$

Simplify each equation.

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

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Simplify each term.

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Apply the distributive property.

$a = 5 - b$

$10 = 5 \cdot 5 + 5 (- b) + b$

$15 = 9 (5 - b) + b$

Multiply $5$ by $5$ .

$a = 5 - b$

$10 = 25 + 5 (- b) + b$

$15 = 9 (5 - b) + b$

Multiply $- 1$ by $5$ .

$a = 5 - b$

$10 = 25 - 5 b + b$

$15 = 9 (5 - b) + b$

$a = 5 - b$

$10 = 25 - 5 b + b$

$15 = 9 (5 - b) + b$

Add $- 5 b$ and $b$ .

$a = 5 - b$

$10 = 25 - 4 b$

$15 = 9 (5 - b) + b$

$a = 5 - b$

$10 = 25 - 4 b$

$15 = 9 (5 - b) + b$

Simplify $9 (5 - b) + b$ .

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Simplify each term.

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Apply the distributive property.

$a = 5 - b$

$10 = 25 - 4 b$

$15 = 9 \cdot 5 + 9 (- b) + b$

Multiply $9$ by $5$ .

$a = 5 - b$

$10 = 25 - 4 b$

$15 = 45 + 9 (- b) + b$

Multiply $- 1$ by $9$ .

$a = 5 - b$

$10 = 25 - 4 b$

$15 = 45 - 9 b + b$

$a = 5 - b$

$10 = 25 - 4 b$

$15 = 45 - 9 b + b$

Add $- 9 b$ and $b$ .

$a = 5 - b$

$10 = 25 - 4 b$

$15 = 45 - 8 b$

$a = 5 - b$

$10 = 25 - 4 b$

$15 = 45 - 8 b$

$a = 5 - b$

$10 = 25 - 4 b$

$15 = 45 - 8 b$

Solve for $b$ in the second equation.

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Rewrite the equation as $25 - 4 b = 10$ .

$a = 5 - b$

$25 - 4 b = 10$

$15 = 45 - 8 b$

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

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Subtract $25$ from both sides of the equation.

$a = 5 - b$

$- 4 b = 10 - 25$

$15 = 45 - 8 b$

Subtract $25$ from $10$ .

$a = 5 - b$

$- 4 b = - 15$

$15 = 45 - 8 b$

$a = 5 - b$

$- 4 b = - 15$

$15 = 45 - 8 b$

Divide each term by $- 4$ and simplify.

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Divide each term in $- 4 b = - 15$ by $- 4$ .

$a = 5 - b$

$\frac{- 4 b}{- 4} = \frac{- 15}{- 4}$

$15 = 45 - 8 b$

Cancel the common factor of $- 4$ .

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Cancel the common factor.

$a = 5 - b$

$\frac{- 4 b}{- 4} = \frac{- 15}{- 4}$

$15 = 45 - 8 b$

Divide $b$ by $1$ .

$a = 5 - b$

$b = \frac{- 15}{- 4}$

$15 = 45 - 8 b$

$a = 5 - b$

$b = \frac{- 15}{- 4}$

$15 = 45 - 8 b$

Dividing two negative values results in a positive value.

$a = 5 - b$

$b = \frac{15}{4}$

$15 = 45 - 8 b$

$a = 5 - b$

$b = \frac{15}{4}$

$15 = 45 - 8 b$

$a = 5 - b$

$b = \frac{15}{4}$

$15 = 45 - 8 b$

Replace all occurrences of $b$ with $\frac{15}{4}$ in each equation.

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

$a = 5 - (\frac{15}{4})$

$b = \frac{15}{4}$

$15 = 45 - 8 b$

Replace all occurrences of $b$ in $15 = 45 - 8 b$ with $\frac{15}{4}$ .

$a = 5 - (\frac{15}{4})$

$b = \frac{15}{4}$

$15 = 45 - 8 (\frac{15}{4})$

$a = 5 - (\frac{15}{4})$

$b = \frac{15}{4}$

$15 = 45 - 8 (\frac{15}{4})$

Simplify.

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

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

$a = 5 \cdot \frac{4}{4} - \frac{15}{4}$

$b = \frac{15}{4}$

$15 = 45 - 8 (\frac{15}{4})$

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

$a = \frac{5 \cdot 4}{4} - \frac{15}{4}$

$b = \frac{15}{4}$

$15 = 45 - 8 (\frac{15}{4})$

Combine the numerators over the common denominator.

$a = \frac{5 \cdot 4 - 15}{4}$

$b = \frac{15}{4}$

$15 = 45 - 8 (\frac{15}{4})$

Simplify the numerator.

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Multiply $5$ by $4$ .

$a = \frac{20 - 15}{4}$

$b = \frac{15}{4}$

$15 = 45 - 8 (\frac{15}{4})$

Subtract $15$ from $20$ .

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

$b = \frac{15}{4}$

$15 = 45 - 8 (\frac{15}{4})$

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

$b = \frac{15}{4}$

$15 = 45 - 8 (\frac{15}{4})$

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

$b = \frac{15}{4}$

$15 = 45 - 8 (\frac{15}{4})$

Simplify $45 - 8 (\frac{15}{4})$ .

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Simplify each term.

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Cancel the common factor of $4$ .

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Factor $4$ out of $- 8$ .

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

$b = \frac{15}{4}$

$15 = 45 + 4 (- 2) (\frac{15}{4})$

Cancel the common factor.

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

$b = \frac{15}{4}$

$15 = 45 + 4 \cdot (- 2 (\frac{15}{4}))$

Rewrite the expression.

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

$b = \frac{15}{4}$

$15 = 45 - 2 \cdot 15$

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

$b = \frac{15}{4}$

$15 = 45 - 2 \cdot 15$

Multiply $- 2$ by $15$ .

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

$b = \frac{15}{4}$

$15 = 45 - 30$

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

$b = \frac{15}{4}$

$15 = 45 - 30$

Subtract $30$ from $45$ .

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

$b = \frac{15}{4}$

$15 = 15$

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

$b = \frac{15}{4}$

$15 = 15$

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

$b = \frac{15}{4}$

$15 = 15$

Since $15 = 15$ , the equation will always be true.

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

$b = \frac{15}{4}$

Always true

Remove any equations from the system that are always true.

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

$b = \frac{15}{4}$

List the solutions to the system of equations.

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

$b = \frac{15}{4}$

Always true

List all of the solutions.

$a = \frac{5}{4} b = \frac{15}{4}$

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

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Calculate the value of $y$ when $a = \frac{5}{4}$ , $b = \frac{15}{4}$ , and $x = 1$ .

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Multiply $\frac{5}{4}$ by $1$ .

$y = \frac{5}{4} + \frac{15}{4}$

Combine the numerators over the common denominator.

$y = \frac{5 + 15}{4}$

Add $5$ and $15$ .

$y = \frac{20}{4}$

Divide $20$ by $4$ .

$y = 5$

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

$5 = 5$

Calculate the value of $y$ when $a = \frac{5}{4}$ , $b = \frac{15}{4}$ , and $x = 5$ .

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

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

$y = \frac{5 \cdot 5}{4} + \frac{15}{4}$

Multiply $5$ by $5$ .

$y = \frac{25}{4} + \frac{15}{4}$

Combine the numerators over the common denominator.

$y = \frac{25 + 15}{4}$

Add $25$ and $15$ .

$y = \frac{40}{4}$

Divide $40$ by $4$ .

$y = 10$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 5$ . This check passes since $y = 10$ and $y = 10$ .

$10 = 10$

Calculate the value of $y$ when $a = \frac{5}{4}$ , $b = \frac{15}{4}$ , and $x = 9$ .

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Multiply $(\frac{5}{4}) (9)$ .

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Combine $\frac{5}{4}$ and $9$ .

$y = \frac{5 \cdot 9}{4} + \frac{15}{4}$

Multiply $5$ by $9$ .

$y = \frac{45}{4} + \frac{15}{4}$

Combine the numerators over the common denominator.

$y = \frac{45 + 15}{4}$

Add $45$ and $15$ .

$y = \frac{60}{4}$

Divide $60$ by $4$ .

$y = 15$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 9$ . This check passes since $y = 15$ and $y = 15$ .

$15 = 15$

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

The function is linear

Since all $y = y$ , the function is linear and follows the form $y = \frac{5 x}{4} + \frac{15}{4}$ .

$y = \frac{5 x}{4} + \frac{15}{4}$

Find the Function Rule table[[x,y],[1,5],[5,10],[9,15]]