Find the Function Rule table[[x,y],[3,15],[8,40],[10,50]]

$\begin{array}{c} x & y \\ 3 & 15 \\ 8 & 40 \\ 10 & 50 \end{array}$

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 $y = a x + b$ .

$15 = a (3) + b 40 = a (8) + b 50 = a (10) + b$

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

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

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

$a (3) + b = 15$

$40 = a (8) + b$

$50 = a (10) + b$

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

$3 a + b = 15$

$40 = a (8) + b$

$50 = a (10) + b$

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

$b = 15 - 3 a$

$40 = a (8) + b$

$50 = a (10) + b$

$b = 15 - 3 a$

$40 = a (8) + b$

$50 = a (10) + b$

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

Replace all occurrences of $b$ in $40 = a (8) + b$ with $15 - 3 a$ .

$40 = a (8) + 15 - 3 a$

$b = 15 - 3 a$

$50 = a (10) + b$

Simplify $40 = a (8) + 15 - 3 a$ .

Simplify the left side.

Remove parentheses.

$40 = a (8) + 15 - 3 a$

$b = 15 - 3 a$

$50 = a (10) + b$

$40 = a (8) + 15 - 3 a$

$b = 15 - 3 a$

$50 = a (10) + b$

Simplify the right side.

Simplify $a (8) + 15 - 3 a$ .

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

$40 = 8 a + 15 - 3 a$

$b = 15 - 3 a$

$50 = a (10) + b$

Subtract $3 a$ from $8 a$ .

$40 = 5 a + 15$

$b = 15 - 3 a$

$50 = a (10) + b$

$40 = 5 a + 15$

$b = 15 - 3 a$

$50 = a (10) + b$

$40 = 5 a + 15$

$b = 15 - 3 a$

$50 = a (10) + b$

$40 = 5 a + 15$

$b = 15 - 3 a$

$50 = a (10) + b$

Replace all occurrences of $b$ in $50 = a (10) + b$ with $15 - 3 a$ .

$50 = a (10) + 15 - 3 a$

$40 = 5 a + 15$

$b = 15 - 3 a$

Simplify $50 = a (10) + 15 - 3 a$ .

Simplify the left side.

Remove parentheses.

$50 = a (10) + 15 - 3 a$

$40 = 5 a + 15$

$b = 15 - 3 a$

$50 = a (10) + 15 - 3 a$

$40 = 5 a + 15$

$b = 15 - 3 a$

Simplify the right side.

Simplify $a (10) + 15 - 3 a$ .

Move $10$ to the left of $a$ .

$50 = 10 a + 15 - 3 a$

$40 = 5 a + 15$

$b = 15 - 3 a$

Subtract $3 a$ from $10 a$ .

$50 = 7 a + 15$

$40 = 5 a + 15$

$b = 15 - 3 a$

$50 = 7 a + 15$

$40 = 5 a + 15$

$b = 15 - 3 a$

$50 = 7 a + 15$

$40 = 5 a + 15$

$b = 15 - 3 a$

$50 = 7 a + 15$

$40 = 5 a + 15$

$b = 15 - 3 a$

$50 = 7 a + 15$

$40 = 5 a + 15$

$b = 15 - 3 a$

Solve for $a$ in $50 = 7 a + 15$ .

Rewrite the equation as $7 a + 15 = 50$ .

$7 a + 15 = 50$

$40 = 5 a + 15$

$b = 15 - 3 a$

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

Subtract $15$ from both sides of the equation.

$7 a = 50 - 15$

$40 = 5 a + 15$

$b = 15 - 3 a$

Subtract $15$ from $50$ .

$7 a = 35$

$40 = 5 a + 15$

$b = 15 - 3 a$

$7 a = 35$

$40 = 5 a + 15$

$b = 15 - 3 a$

Divide each term in $7 a = 35$ by $7$ and simplify.

Divide each term in $7 a = 35$ by $7$ .

$\frac{7 a}{7} = \frac{35}{7}$

$40 = 5 a + 15$

$b = 15 - 3 a$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 a}{7} = \frac{35}{7}$

$40 = 5 a + 15$

$b = 15 - 3 a$

Divide $a$ by $1$ .

$a = \frac{35}{7}$

$40 = 5 a + 15$

$b = 15 - 3 a$

$a = \frac{35}{7}$

$40 = 5 a + 15$

$b = 15 - 3 a$

$a = \frac{35}{7}$

$40 = 5 a + 15$

$b = 15 - 3 a$

Simplify the right side.

Divide $35$ by $7$ .

$a = 5$

$40 = 5 a + 15$

$b = 15 - 3 a$

$a = 5$

$40 = 5 a + 15$

$b = 15 - 3 a$

$a = 5$

$40 = 5 a + 15$

$b = 15 - 3 a$

$a = 5$

$40 = 5 a + 15$

$b = 15 - 3 a$

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

Replace all occurrences of $a$ in $40 = 5 a + 15$ with $5$ .

$40 = 5 (5) + 15$

$a = 5$

$b = 15 - 3 a$

Simplify the right side.

Simplify $5 (5) + 15$ .

Multiply $5$ by $5$ .

$40 = 25 + 15$

$a = 5$

$b = 15 - 3 a$

Add $25$ and $15$ .

$40 = 40$

$a = 5$

$b = 15 - 3 a$

$40 = 40$

$a = 5$

$b = 15 - 3 a$

$40 = 40$

$a = 5$

$b = 15 - 3 a$

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

$b = 15 - 3 \cdot 5$

$40 = 40$

$a = 5$

Simplify the right side.

Simplify $15 - 3 \cdot 5$ .

Multiply $- 3$ by $5$ .

$b = 15 - 15$

$40 = 40$

$a = 5$

Subtract $15$ from $15$ .

$b = 0$

$40 = 40$

$a = 5$

$b = 0$

$40 = 40$

$a = 5$

$b = 0$

$40 = 40$

$a = 5$

$b = 0$

$40 = 40$

$a = 5$

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 $x = 8$ .

Multiply $5$ by $8$ .

$y = 40 + 0$

Add $40$ and $0$ .

$y = 40$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 8$ . This check passes since $y = 40$ and $y = 40$ .

$40 = 40$

Calculate the value of $y$ when $a = 5$ , $b = 0$ , and $x = 10$ .

Multiply $5$ by $10$ .

$y = 50 + 0$

Add $50$ and $0$ .

$y = 50$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 10$ . This check passes since $y = 50$ and $y = 50$ .

$50 = 50$

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],[8,40],[10,50]]

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