Pre-Algebra Examples

$\begin{array}{c} x & y \\ 1 & 6 \\ 2 & 4 \\ 3 & 2 \\ 4 & 0 \end{array}$

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 $y = a x + b$ .

$6 = a (1) + b 4 = a (2) + b 2 = a (3) + b 0 = a (4) + b$

Step 1.3

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

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Step 1.3.1

Solve for $a$ in $6 = a + b$ .

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Step 1.3.1.1

Rewrite the equation as $a + b = 6$ .

$a + b = 6$

$4 = a (2) + b$

$2 = a (3) + b$

$0 = a (4) + b$

Step 1.3.1.2

Subtract $b$ from both sides of the equation.

$a = 6 - b$

$4 = a (2) + b$

$2 = a (3) + b$

$0 = a (4) + b$

$a = 6 - b$

$4 = a (2) + b$

$2 = a (3) + b$

$0 = a (4) + b$

Step 1.3.2

Replace all occurrences of $a$ with $6 - b$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $a$ in $4 = a (2) + b$ with $6 - b$ .

$4 = (6 - b) (2) + b$

$a = 6 - b$

$2 = a (3) + b$

$0 = a (4) + b$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $(6 - b) (2) + b$ .

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Step 1.3.2.2.1.1

Simplify each term.

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Step 1.3.2.2.1.1.1

Apply the distributive property.

$4 = 6 \cdot 2 - b \cdot 2 + b$

$a = 6 - b$

$2 = a (3) + b$

$0 = a (4) + b$

Step 1.3.2.2.1.1.2

Multiply $6$ by $2$ .

$4 = 12 - b \cdot 2 + b$

$a = 6 - b$

$2 = a (3) + b$

$0 = a (4) + b$

Step 1.3.2.2.1.1.3

Multiply $2$ by $- 1$ .

$4 = 12 - 2 b + b$

$a = 6 - b$

$2 = a (3) + b$

$0 = a (4) + b$

$4 = 12 - 2 b + b$

$a = 6 - b$

$2 = a (3) + b$

$0 = a (4) + b$

Step 1.3.2.2.1.2

Add $- 2 b$ and $b$ .

$4 = 12 - b$

$a = 6 - b$

$2 = a (3) + b$

$0 = a (4) + b$

$4 = 12 - b$

$a = 6 - b$

$2 = a (3) + b$

$0 = a (4) + b$

$4 = 12 - b$

$a = 6 - b$

$2 = a (3) + b$

$0 = a (4) + b$

Step 1.3.2.3

Replace all occurrences of $a$ in $2 = a (3) + b$ with $6 - b$ .

$2 = (6 - b) (3) + b$

$4 = 12 - b$

$a = 6 - b$

$0 = a (4) + b$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

Simplify $(6 - b) (3) + b$ .

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Step 1.3.2.4.1.1

Simplify each term.

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Step 1.3.2.4.1.1.1

Apply the distributive property.

$2 = 6 \cdot 3 - b \cdot 3 + b$

$4 = 12 - b$

$a = 6 - b$

$0 = a (4) + b$

Step 1.3.2.4.1.1.2

Multiply $6$ by $3$ .

$2 = 18 - b \cdot 3 + b$

$4 = 12 - b$

$a = 6 - b$

$0 = a (4) + b$

Step 1.3.2.4.1.1.3

Multiply $3$ by $- 1$ .

$2 = 18 - 3 b + b$

$4 = 12 - b$

$a = 6 - b$

$0 = a (4) + b$

$2 = 18 - 3 b + b$

$4 = 12 - b$

$a = 6 - b$

$0 = a (4) + b$

Step 1.3.2.4.1.2

Add $- 3 b$ and $b$ .

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$0 = a (4) + b$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$0 = a (4) + b$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$0 = a (4) + b$

Step 1.3.2.5

Replace all occurrences of $a$ in $0 = a (4) + b$ with $6 - b$ .

$0 = (6 - b) (4) + b$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.2.6

Simplify the right side.

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Step 1.3.2.6.1

Simplify $(6 - b) (4) + b$ .

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Step 1.3.2.6.1.1

Simplify each term.

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Step 1.3.2.6.1.1.1

Apply the distributive property.

$0 = 6 \cdot 4 - b \cdot 4 + b$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.2.6.1.1.2

Multiply $6$ by $4$ .

$0 = 24 - b \cdot 4 + b$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.2.6.1.1.3

Multiply $4$ by $- 1$ .

$0 = 24 - 4 b + b$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$0 = 24 - 4 b + b$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.2.6.1.2

Add $- 4 b$ and $b$ .

$0 = 24 - 3 b$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$0 = 24 - 3 b$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$0 = 24 - 3 b$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$0 = 24 - 3 b$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.3

Solve for $b$ in $0 = 24 - 3 b$ .

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Step 1.3.3.1

Rewrite the equation as $24 - 3 b = 0$ .

$24 - 3 b = 0$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.3.2

Subtract $24$ from both sides of the equation.

$- 3 b = - 24$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.3.3

Divide each term in $- 3 b = - 24$ by $- 3$ and simplify.

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Step 1.3.3.3.1

Divide each term in $- 3 b = - 24$ by $- 3$ .

$\frac{- 3 b}{- 3} = \frac{- 24}{- 3}$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

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 $- 3$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{- 3 b}{- 3} = \frac{- 24}{- 3}$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.3.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{- 24}{- 3}$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$b = \frac{- 24}{- 3}$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$b = \frac{- 24}{- 3}$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide $- 24$ by $- 3$ .

$b = 8$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$b = 8$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$b = 8$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

$b = 8$

$2 = 18 - 2 b$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.4

Replace all occurrences of $b$ with $8$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $b$ in $2 = 18 - 2 b$ with $8$ .

$2 = 18 - 2 \cdot 8$

$b = 8$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $18 - 2 \cdot 8$ .

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Step 1.3.4.2.1.1

Multiply $- 2$ by $8$ .

$2 = 18 - 16$

$b = 8$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.4.2.1.2

Subtract $16$ from $18$ .

$2 = 2$

$b = 8$

$4 = 12 - b$

$a = 6 - b$

$2 = 2$

$b = 8$

$4 = 12 - b$

$a = 6 - b$

$2 = 2$

$b = 8$

$4 = 12 - b$

$a = 6 - b$

Step 1.3.4.3

Replace all occurrences of $b$ in $4 = 12 - b$ with $8$ .

$4 = 12 - (8)$

$2 = 2$

$b = 8$

$a = 6 - b$

Step 1.3.4.4

Simplify the right side.

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Step 1.3.4.4.1

Simplify $12 - (8)$ .

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Step 1.3.4.4.1.1

Multiply $- 1$ by $8$ .

$4 = 12 - 8$

$2 = 2$

$b = 8$

$a = 6 - b$

Step 1.3.4.4.1.2

Subtract $8$ from $12$ .

$4 = 4$

$2 = 2$

$b = 8$

$a = 6 - b$

$4 = 4$

$2 = 2$

$b = 8$

$a = 6 - b$

$4 = 4$

$2 = 2$

$b = 8$

$a = 6 - b$

Step 1.3.4.5

Replace all occurrences of $b$ in $a = 6 - b$ with $8$ .

$a = 6 - (8)$

$4 = 4$

$2 = 2$

$b = 8$

Step 1.3.4.6

Simplify the right side.

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Step 1.3.4.6.1

Simplify $6 - (8)$ .

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Step 1.3.4.6.1.1

Multiply $- 1$ by $8$ .

$a = 6 - 8$

$4 = 4$

$2 = 2$

$b = 8$

Step 1.3.4.6.1.2

Subtract $8$ from $6$ .

$a = - 2$

$4 = 4$

$2 = 2$

$b = 8$

$a = - 2$

$4 = 4$

$2 = 2$

$b = 8$

$a = - 2$

$4 = 4$

$2 = 2$

$b = 8$

$a = - 2$

$4 = 4$

$2 = 2$

$b = 8$

Step 1.3.5

Remove any equations from the system that are always true.

$a = - 2$

$b = 8$

Step 1.3.6

List all of the solutions.

$a = - 2, b = 8$

Step 1.4

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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Step 1.4.1

Calculate the value of $y$ when $a = - 2$ , $b = 8$ , and $x = 1$ .

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Step 1.4.1.1

Multiply $- 2$ by $1$ .

$y = - 2 + 8$

Step 1.4.1.2

Add $- 2$ and $8$ .

$y = 6$

Step 1.4.2

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 6$ and $y = 6$ .

$6 = 6$

Step 1.4.3

Calculate the value of $y$ when $a = - 2$ , $b = 8$ , and $x = 2$ .

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Step 1.4.3.1

Multiply $- 2$ by $2$ .

$y = - 4 + 8$

Step 1.4.3.2

Add $- 4$ and $8$ .

$y = 4$

Step 1.4.4

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 4$ and $y = 4$ .

$4 = 4$

Step 1.4.5

Calculate the value of $y$ when $a = - 2$ , $b = 8$ , and $x = 3$ .

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Step 1.4.5.1

Multiply $- 2$ by $3$ .

$y = - 6 + 8$

Step 1.4.5.2

Add $- 6$ and $8$ .

$y = 2$

Step 1.4.6

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 2$ and $y = 2$ .

$2 = 2$

Step 1.4.7

Calculate the value of $y$ when $a = - 2$ , $b = 8$ , and $x = 4$ .

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Step 1.4.7.1

Multiply $- 2$ by $4$ .

$y = - 8 + 8$

Step 1.4.7.2

Add $- 8$ and $8$ .

$y = 0$

Step 1.4.8

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = 0$ and $y = 0$ .

$0 = 0$

Step 1.4.9

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 $y = - 2 x + 8$ .

$y = - 2 x + 8$