Find the Function Rule table[[x,y],[3,15],[5,25],[7,35]]

$\begin{array}{c} x & y \\ 3 & 15 \\ 5 & 25 \\ 7 & 35 \end{array}$

Step 1

Check if the function rule is linear.

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 $y = a x + b$ .

$15 = a (3) + b 25 = a (5) + b 35 = a (7) + b$

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

Solve for $b$ in $15 = a (3) + b$ .

Rewrite the equation as $a (3) + b = 15$ .

$a (3) + b = 15$

$25 = a (5) + b$

$35 = a (7) + b$

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

$3 a + b = 15$

$25 = a (5) + b$

$35 = a (7) + b$

Subtract $3 a$ from both sides of the equation.

$b = 15 - 3 a$

$25 = a (5) + b$

$35 = a (7) + b$

$b = 15 - 3 a$

$25 = a (5) + b$

$35 = a (7) + b$

Replace all occurrences of $b$ with $15 - 3 a$ in each equation.

Replace all occurrences of $b$ in $25 = a (5) + b$ with $15 - 3 a$ .

$25 = a (5) + 15 - 3 a$

$b = 15 - 3 a$

$35 = a (7) + b$

Simplify $25 = a (5) + 15 - 3 a$ .

Simplify the left side.

Remove parentheses.

$25 = a (5) + 15 - 3 a$

$b = 15 - 3 a$

$35 = a (7) + b$

$25 = a (5) + 15 - 3 a$

$b = 15 - 3 a$

$35 = a (7) + b$

Simplify the right side.

Simplify $a (5) + 15 - 3 a$ .

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

$25 = 5 a + 15 - 3 a$

$b = 15 - 3 a$

$35 = a (7) + b$

Subtract $3 a$ from $5 a$ .

$25 = 2 a + 15$

$b = 15 - 3 a$

$35 = a (7) + b$

$25 = 2 a + 15$

$b = 15 - 3 a$

$35 = a (7) + b$

$25 = 2 a + 15$

$b = 15 - 3 a$

$35 = a (7) + b$

$25 = 2 a + 15$

$b = 15 - 3 a$

$35 = a (7) + b$

Replace all occurrences of $b$ in $35 = a (7) + b$ with $15 - 3 a$ .

$35 = a (7) + 15 - 3 a$

$25 = 2 a + 15$

$b = 15 - 3 a$

Simplify $35 = a (7) + 15 - 3 a$ .

Simplify the left side.

Remove parentheses.

$35 = a (7) + 15 - 3 a$

$25 = 2 a + 15$

$b = 15 - 3 a$

$35 = a (7) + 15 - 3 a$

$25 = 2 a + 15$

$b = 15 - 3 a$

Simplify the right side.

Simplify $a (7) + 15 - 3 a$ .

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

$35 = 7 a + 15 - 3 a$

$25 = 2 a + 15$

$b = 15 - 3 a$

Subtract $3 a$ from $7 a$ .

$35 = 4 a + 15$

$25 = 2 a + 15$

$b = 15 - 3 a$

$35 = 4 a + 15$

$25 = 2 a + 15$

$b = 15 - 3 a$

$35 = 4 a + 15$

$25 = 2 a + 15$

$b = 15 - 3 a$

$35 = 4 a + 15$

$25 = 2 a + 15$

$b = 15 - 3 a$

$35 = 4 a + 15$

$25 = 2 a + 15$

$b = 15 - 3 a$

Solve for $a$ in $35 = 4 a + 15$ .

Rewrite the equation as $4 a + 15 = 35$ .

$4 a + 15 = 35$

$25 = 2 a + 15$

$b = 15 - 3 a$

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

Subtract $15$ from both sides of the equation.

$4 a = 35 - 15$

$25 = 2 a + 15$

$b = 15 - 3 a$

Subtract $15$ from $35$ .

$4 a = 20$

$25 = 2 a + 15$

$b = 15 - 3 a$

$4 a = 20$

$25 = 2 a + 15$

$b = 15 - 3 a$

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

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

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

$25 = 2 a + 15$

$b = 15 - 3 a$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

$25 = 2 a + 15$

$b = 15 - 3 a$

Divide $a$ by $1$ .

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

$25 = 2 a + 15$

$b = 15 - 3 a$

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

$25 = 2 a + 15$

$b = 15 - 3 a$

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

$25 = 2 a + 15$

$b = 15 - 3 a$

Simplify the right side.

Divide $20$ by $4$ .

$a = 5$

$25 = 2 a + 15$

$b = 15 - 3 a$

$a = 5$

$25 = 2 a + 15$

$b = 15 - 3 a$

$a = 5$

$25 = 2 a + 15$

$b = 15 - 3 a$

$a = 5$

$25 = 2 a + 15$

$b = 15 - 3 a$

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

Replace all occurrences of $a$ in $25 = 2 a + 15$ with $5$ .

$25 = 2 (5) + 15$

$a = 5$

$b = 15 - 3 a$

Simplify the right side.

Simplify $2 (5) + 15$ .

Multiply $2$ by $5$ .

$25 = 10 + 15$

$a = 5$

$b = 15 - 3 a$

Add $10$ and $15$ .

$25 = 25$

$a = 5$

$b = 15 - 3 a$

$25 = 25$

$a = 5$

$b = 15 - 3 a$

$25 = 25$

$a = 5$

$b = 15 - 3 a$

Replace all occurrences of $a$ in $b = 15 - 3 a$ with $5$ .

$b = 15 - 3 \cdot 5$

$25 = 25$

$a = 5$

Simplify the right side.

Simplify $15 - 3 \cdot 5$ .

Multiply $- 3$ by $5$ .

$b = 15 - 15$

$25 = 25$

$a = 5$

Subtract $15$ from $15$ .

$b = 0$

$25 = 25$

$a = 5$

$b = 0$

$25 = 25$

$a = 5$

$b = 0$

$25 = 25$

$a = 5$

$b = 0$

$25 = 25$

$a = 5$

Remove any equations from the system that are always true.

$b = 0$

$a = 5$

List all of the solutions.

$b = 0, a = 5$

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

Calculate the value of $y$ when $a = 5$ , $b = 0$ , and $x = 3$ .

Multiply $5$ by $3$ .

$y = 15 + 0$

Add $15$ and $0$ .

$y = 15$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 15$ and $y = 15$ .

$15 = 15$

Calculate the value of $y$ when $a = 5$ , $b = 0$ , and $x = 5$ .

Multiply $5$ by $5$ .

$y = 25 + 0$

Add $25$ and $0$ .

$y = 25$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 5$ . This check passes since $y = 25$ and $y = 25$ .

$25 = 25$

Calculate the value of $y$ when $a = 5$ , $b = 0$ , and $x = 7$ .

Multiply $5$ by $7$ .

$y = 35 + 0$

Add $35$ and $0$ .

$y = 35$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 7$ . This check passes since $y = 35$ and $y = 35$ .

$35 = 35$

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

The function is linear

Step 2

Since all $y = y$ , the function is linear and follows the form $y = 5 x$ .

$y = 5 x$

Find the Function Rule table[[x,y],[3,15],[5,25],[7,35]]

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