Find the Function Rule table[[x,y],[1,5],[2,10],[3,15]]

$\begin{array}{c} x & y \\ 1 & 5 \\ 2 & 10 \\ 3 & 15 \end{array}$

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 $y = a x + b$ .

$5 = a (1) + b 10 = a (2) + b 15 = a (3) + b$

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

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

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

$a + b = 5$

$10 = a (2) + b$

$15 = a (3) + b$

Subtract $b$ from both sides of the equation.

$a = 5 - b$

$10 = a (2) + b$

$15 = a (3) + b$

$a = 5 - b$

$10 = a (2) + b$

$15 = a (3) + b$

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

Replace all occurrences of $a$ in $10 = a (2) + b$ with $5 - b$ .

$10 = (5 - b) (2) + b$

$a = 5 - b$

$15 = a (3) + b$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

$10 = 5 \cdot 2 - b \cdot 2 + b$

$a = 5 - b$

$15 = a (3) + b$

Multiply $5$ by $2$ .

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

$a = 5 - b$

$15 = a (3) + b$

Multiply $2$ by $- 1$ .

$10 = 10 - 2 b + b$

$a = 5 - b$

$15 = a (3) + b$

$10 = 10 - 2 b + b$

$a = 5 - b$

$15 = a (3) + b$

Add $- 2 b$ and $b$ .

$10 = 10 - b$

$a = 5 - b$

$15 = a (3) + b$

$10 = 10 - b$

$a = 5 - b$

$15 = a (3) + b$

$10 = 10 - b$

$a = 5 - b$

$15 = a (3) + b$

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

$15 = (5 - b) (3) + b$

$10 = 10 - b$

$a = 5 - b$

Simplify the right side.

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

Simplify each term.

Apply the distributive property.

$15 = 5 \cdot 3 - b \cdot 3 + b$

$10 = 10 - b$

$a = 5 - b$

Multiply $5$ by $3$ .

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

$10 = 10 - b$

$a = 5 - b$

Multiply $3$ by $- 1$ .

$15 = 15 - 3 b + b$

$10 = 10 - b$

$a = 5 - b$

$15 = 15 - 3 b + b$

$10 = 10 - b$

$a = 5 - b$

Add $- 3 b$ and $b$ .

$15 = 15 - 2 b$

$10 = 10 - b$

$a = 5 - b$

$15 = 15 - 2 b$

$10 = 10 - b$

$a = 5 - b$

$15 = 15 - 2 b$

$10 = 10 - b$

$a = 5 - b$

$15 = 15 - 2 b$

$10 = 10 - b$

$a = 5 - b$

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

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

$15 - 2 b = 15$

$10 = 10 - b$

$a = 5 - b$

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

Subtract $15$ from both sides of the equation.

$- 2 b = 15 - 15$

$10 = 10 - b$

$a = 5 - b$

Subtract $15$ from $15$ .

$- 2 b = 0$

$10 = 10 - b$

$a = 5 - b$

$- 2 b = 0$

$10 = 10 - b$

$a = 5 - b$

Divide each term in $- 2 b = 0$ by $- 2$ and simplify.

Divide each term in $- 2 b = 0$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{0}{- 2}$

$10 = 10 - b$

$a = 5 - b$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{0}{- 2}$

$10 = 10 - b$

$a = 5 - b$

Divide $b$ by $1$ .

$b = \frac{0}{- 2}$

$10 = 10 - b$

$a = 5 - b$

$b = \frac{0}{- 2}$

$10 = 10 - b$

$a = 5 - b$

$b = \frac{0}{- 2}$

$10 = 10 - b$

$a = 5 - b$

Simplify the right side.

Divide $0$ by $- 2$ .

$b = 0$

$10 = 10 - b$

$a = 5 - b$

$b = 0$

$10 = 10 - b$

$a = 5 - b$

$b = 0$

$10 = 10 - b$

$a = 5 - b$

$b = 0$

$10 = 10 - b$

$a = 5 - b$

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

Replace all occurrences of $b$ in $10 = 10 - b$ with $0$ .

$10 = 10 - (0)$

$b = 0$

$a = 5 - b$

Simplify the right side.

Subtract $0$ from $10$ .

$10 = 10$

$b = 0$

$a = 5 - b$

$10 = 10$

$b = 0$

$a = 5 - b$

Replace all occurrences of $b$ in $a = 5 - b$ with $0$ .

$a = 5 - (0)$

$10 = 10$

$b = 0$

Simplify the right side.

Subtract $0$ from $5$ .

$a = 5$

$10 = 10$

$b = 0$

$a = 5$

$10 = 10$

$b = 0$

$a = 5$

$10 = 10$

$b = 0$

Remove any equations from the system that are always true.

$a = 5$

$b = 0$

List all of the solutions.

$a = 5, b = 0$

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 $x = 1$ .

Multiply $5$ by $1$ .

$y = 5 + 0$

Add $5$ and $0$ .

$y = 5$

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 $a = 5$ , $b = 0$ , and $x = 2$ .

Multiply $5$ by $2$ .

$y = 10 + 0$

Add $10$ and $0$ .

$y = 10$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 10$ and $y = 10$ .

$10 = 10$

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 $y = 15$ .

$15 = 15$

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],[1,5],[2,10],[3,15]]

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