Examples

$\begin{array}{c} x & q (x) \\ 4 & 10 \\ 7 & 16 \\ 2 & 6 \end{array}$

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

$q (x) = a x + b$ .

$10 = a (4) + b 16 = a (7) + b 6 = a (2) + b$

Step 1.3

Calculate the values of

$a$ and

$b$ .

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Step 1.3.1

Solve for

$b$ in

$10 = a (4) + b$ .

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Step 1.3.1.1

Rewrite the equation as

$a (4) + b = 10$ .

$a (4) + b = 10$

$16 = a (7) + b$

$6 = a (2) + b$

Step 1.3.1.2

Move

$4$ to the left of

$a$ .

$4 a + b = 10$

$16 = a (7) + b$

$6 = a (2) + b$

Step 1.3.1.3

Subtract

$4 a$ from both sides of the equation.

$b = 10 - 4 a$

$16 = a (7) + b$

$6 = a (2) + b$

$b = 10 - 4 a$

$16 = a (7) + b$

$6 = a (2) + b$

Step 1.3.2

Replace all occurrences of

$b$ with

$10 - 4 a$ in each equation.

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Step 1.3.2.1

Replace all occurrences of

$b$ in

$16 = a (7) + b$ with

$10 - 4 a$ .

$16 = a (7) + 10 - 4 a$

$b = 10 - 4 a$

$6 = a (2) + b$

Step 1.3.2.2

Simplify

$16 = a (7) + 10 - 4 a$ .

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Step 1.3.2.2.1

Simplify the left side.

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Step 1.3.2.2.1.1

Remove parentheses.

$16 = a (7) + 10 - 4 a$

$b = 10 - 4 a$

$6 = a (2) + b$

$16 = a (7) + 10 - 4 a$

$b = 10 - 4 a$

$6 = a (2) + b$

Step 1.3.2.2.2

Simplify the right side.

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Step 1.3.2.2.2.1

Simplify

$a (7) + 10 - 4 a$ .

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Step 1.3.2.2.2.1.1

Move

$7$ to the left of

$a$ .

$16 = 7 a + 10 - 4 a$

$b = 10 - 4 a$

$6 = a (2) + b$

Step 1.3.2.2.2.1.2

Subtract

$4 a$ from

$7 a$ .

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = a (2) + b$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = a (2) + b$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = a (2) + b$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = a (2) + b$

Step 1.3.2.3

Replace all occurrences of

$b$ in

$6 = a (2) + b$ with

$10 - 4 a$ .

$6 = a (2) + 10 - 4 a$

$16 = 3 a + 10$

$b = 10 - 4 a$

Step 1.3.2.4

Simplify

$6 = a (2) + 10 - 4 a$ .

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Step 1.3.2.4.1

Simplify the left side.

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Step 1.3.2.4.1.1

Remove parentheses.

$6 = a (2) + 10 - 4 a$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = a (2) + 10 - 4 a$

$16 = 3 a + 10$

$b = 10 - 4 a$

Step 1.3.2.4.2

Simplify the right side.

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Step 1.3.2.4.2.1

Simplify

$a (2) + 10 - 4 a$ .

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Step 1.3.2.4.2.1.1

Move

$2$ to the left of

$a$ .

$6 = 2 a + 10 - 4 a$

$16 = 3 a + 10$

$b = 10 - 4 a$

Step 1.3.2.4.2.1.2

Subtract

$4 a$ from

$2 a$ .

$6 = - 2 a + 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = - 2 a + 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = - 2 a + 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = - 2 a + 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

$6 = - 2 a + 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

Step 1.3.3

Solve for

$a$ in

$6 = - 2 a + 10$ .

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Step 1.3.3.1

Rewrite the equation as

$- 2 a + 10 = 6$ .

$- 2 a + 10 = 6$

$16 = 3 a + 10$

$b = 10 - 4 a$

Step 1.3.3.2

Move all terms not containing

$a$ to the right side of the equation.

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Step 1.3.3.2.1

Subtract

$10$ from both sides of the equation.

$- 2 a = 6 - 10$

$16 = 3 a + 10$

$b = 10 - 4 a$

Step 1.3.3.2.2

Subtract

$10$ from

$6$ .

$- 2 a = - 4$

$16 = 3 a + 10$

$b = 10 - 4 a$

$- 2 a = - 4$

$16 = 3 a + 10$

$b = 10 - 4 a$

Step 1.3.3.3

Divide each term in

$- 2 a = - 4$ by

$- 2$ and simplify.

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Step 1.3.3.3.1

Divide each term in

$- 2 a = - 4$ by

$- 2$ .

$\frac{- 2 a}{- 2} = \frac{- 4}{- 2}$

$16 = 3 a + 10$

$b = 10 - 4 a$

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.

$\frac{- 2 a}{- 2} = \frac{- 4}{- 2}$

$16 = 3 a + 10$

$b = 10 - 4 a$

Step 1.3.3.3.2.1.2

Divide

$a$ by

$1$ .

$a = \frac{- 4}{- 2}$

$16 = 3 a + 10$

$b = 10 - 4 a$

$a = \frac{- 4}{- 2}$

$16 = 3 a + 10$

$b = 10 - 4 a$

$a = \frac{- 4}{- 2}$

$16 = 3 a + 10$

$b = 10 - 4 a$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide

$- 4$ by

$- 2$ .

$a = 2$

$16 = 3 a + 10$

$b = 10 - 4 a$

$a = 2$

$16 = 3 a + 10$

$b = 10 - 4 a$

$a = 2$

$16 = 3 a + 10$

$b = 10 - 4 a$

$a = 2$

$16 = 3 a + 10$

$b = 10 - 4 a$

Step 1.3.4

Replace all occurrences of

$a$ with

$2$ in each equation.

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Step 1.3.4.1

Replace all occurrences of

$a$ in

$16 = 3 a + 10$ with

$2$ .

$16 = 3 (2) + 10$

$a = 2$

$b = 10 - 4 a$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify

$3 (2) + 10$ .

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Step 1.3.4.2.1.1

Multiply

$3$ by

$2$ .

$16 = 6 + 10$

$a = 2$

$b = 10 - 4 a$

Step 1.3.4.2.1.2

Add

$6$ and

$10$ .

$16 = 16$

$a = 2$

$b = 10 - 4 a$

$16 = 16$

$a = 2$

$b = 10 - 4 a$

$16 = 16$

$a = 2$

$b = 10 - 4 a$

Step 1.3.4.3

Replace all occurrences of

$a$ in

$b = 10 - 4 a$ with

$2$ .

$b = 10 - 4 \cdot 2$

$16 = 16$

$a = 2$

Step 1.3.4.4

Simplify the right side.

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Step 1.3.4.4.1

Simplify

$10 - 4 \cdot 2$ .

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Step 1.3.4.4.1.1

Multiply

$- 4$ by

$2$ .

$b = 10 - 8$

$16 = 16$

$a = 2$

Step 1.3.4.4.1.2

Subtract

$8$ from

$10$ .

$b = 2$

$16 = 16$

$a = 2$

$b = 2$

$16 = 16$

$a = 2$

$b = 2$

$16 = 16$

$a = 2$

$b = 2$

$16 = 16$

$a = 2$

Step 1.3.5

Remove any equations from the system that are always true.

$b = 2$

$a = 2$

Step 1.3.6

List all of the solutions.

$b = 2, a = 2$

Step 1.4

Calculate the value of

$y$ using each

$x$ value in the relation and compare this value to the given

$q (x)$ value in the relation.

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Step 1.4.1

Calculate the value of

$y$ when

$a = 2$ ,

$b = 2$ , and

$x = 4$ .

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Step 1.4.1.1

Multiply

$2$ by

$4$ .

$y = 8 + 2$

Step 1.4.1.2

Add

$8$ and

$2$ .

$y = 10$

Step 1.4.2

If the table has a linear function rule,

$y = q (x)$ for the corresponding

$x$ value,

$x = 4$ . This check passes since

$y = 10$ and

$q (x) = 10$ .

$10 = 10$

Step 1.4.3

Calculate the value of

$y$ when

$a = 2$ ,

$b = 2$ , and

$x = 7$ .

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Step 1.4.3.1

Multiply

$2$ by

$7$ .

$y = 14 + 2$

Step 1.4.3.2

Add

$14$ and

$2$ .

$y = 16$

Step 1.4.4

If the table has a linear function rule,

$y = q (x)$ for the corresponding

$x$ value,

$x = 7$ . This check passes since

$y = 16$ and

$q (x) = 16$ .

$16 = 16$

Step 1.4.5

Calculate the value of

$y$ when

$a = 2$ ,

$b = 2$ , and

$x = 2$ .

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Step 1.4.5.1

Multiply

$2$ by

$2$ .

$y = 4 + 2$

Step 1.4.5.2

Add

$4$ and

$2$ .

$y = 6$

Step 1.4.6

If the table has a linear function rule,

$y = q (x)$ for the corresponding

$x$ value,

$x = 2$ . This check passes since

$y = 6$ and

$q (x) = 6$ .

$6 = 6$

Step 1.4.7

Since

$y = q (x)$ for the corresponding

$x$ values, the function is linear.

The function is linear

Step 2

Since all

$y = q (x)$ , the function is linear and follows the form

$y = 2 x + 2$ .

$y = 2 x + 2$

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