Pre-Algebra Examples

$\begin{array}{c} x & y \\ 1 & 3 \\ 2 & 7 \\ 3 & 11 \end{array}$

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 $y = a x + b$ .

$3 = a (1) + b 7 = a (2) + b 11 = a (3) + b$

Step 1.3

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

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Step 1.3.1

Solve for $a$ in $3 = a + b$ .

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Step 1.3.1.1

Rewrite the equation as $a + b = 3$ .

$a + b = 3$

$7 = a (2) + b$

$11 = a (3) + b$

Step 1.3.1.2

Subtract $b$ from both sides of the equation.

$a = 3 - b$

$7 = a (2) + b$

$11 = a (3) + b$

$a = 3 - b$

$7 = a (2) + b$

$11 = a (3) + b$

Step 1.3.2

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

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Step 1.3.2.1

Replace all occurrences of $a$ in $7 = a (2) + b$ with $3 - b$ .

$7 = (3 - b) (2) + b$

$a = 3 - b$

$11 = a (3) + b$

Step 1.3.2.2

Simplify the right side.

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Step 1.3.2.2.1

Simplify $(3 - b) (2) + b$ .

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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.

$7 = 3 \cdot 2 - b \cdot 2 + b$

$a = 3 - b$

$11 = a (3) + b$

Step 1.3.2.2.1.1.2

Multiply $3$ by $2$ .

$7 = 6 - b \cdot 2 + b$

$a = 3 - b$

$11 = a (3) + b$

Step 1.3.2.2.1.1.3

Multiply $2$ by $- 1$ .

$7 = 6 - 2 b + b$

$a = 3 - b$

$11 = a (3) + b$

$7 = 6 - 2 b + b$

$a = 3 - b$

$11 = a (3) + b$

Step 1.3.2.2.1.2

Add $- 2 b$ and $b$ .

$7 = 6 - b$

$a = 3 - b$

$11 = a (3) + b$

$7 = 6 - b$

$a = 3 - b$

$11 = a (3) + b$

$7 = 6 - b$

$a = 3 - b$

$11 = a (3) + b$

Step 1.3.2.3

Replace all occurrences of $a$ in $11 = a (3) + b$ with $3 - b$ .

$11 = (3 - b) (3) + b$

$7 = 6 - b$

$a = 3 - b$

Step 1.3.2.4

Simplify the right side.

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Step 1.3.2.4.1

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

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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.

$11 = 3 \cdot 3 - b \cdot 3 + b$

$7 = 6 - b$

$a = 3 - b$

Step 1.3.2.4.1.1.2

Multiply $3$ by $3$ .

$11 = 9 - b \cdot 3 + b$

$7 = 6 - b$

$a = 3 - b$

Step 1.3.2.4.1.1.3

Multiply $3$ by $- 1$ .

$11 = 9 - 3 b + b$

$7 = 6 - b$

$a = 3 - b$

$11 = 9 - 3 b + b$

$7 = 6 - b$

$a = 3 - b$

Step 1.3.2.4.1.2

Add $- 3 b$ and $b$ .

$11 = 9 - 2 b$

$7 = 6 - b$

$a = 3 - b$

$11 = 9 - 2 b$

$7 = 6 - b$

$a = 3 - b$

$11 = 9 - 2 b$

$7 = 6 - b$

$a = 3 - b$

$11 = 9 - 2 b$

$7 = 6 - b$

$a = 3 - b$

Step 1.3.3

Solve for $b$ in $11 = 9 - 2 b$ .

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Step 1.3.3.1

Rewrite the equation as $9 - 2 b = 11$ .

$9 - 2 b = 11$

$7 = 6 - b$

$a = 3 - b$

Step 1.3.3.2

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

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Step 1.3.3.2.1

Subtract $9$ from both sides of the equation.

$- 2 b = 11 - 9$

$7 = 6 - b$

$a = 3 - b$

Step 1.3.3.2.2

Subtract $9$ from $11$ .

$- 2 b = 2$

$7 = 6 - b$

$a = 3 - b$

$- 2 b = 2$

$7 = 6 - b$

$a = 3 - b$

Step 1.3.3.3

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

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Step 1.3.3.3.1

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

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

$7 = 6 - b$

$a = 3 - b$

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 $- 2$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

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

$7 = 6 - b$

$a = 3 - b$

Step 1.3.3.3.2.1.2

Divide $b$ by $1$ .

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

$7 = 6 - b$

$a = 3 - b$

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

$7 = 6 - b$

$a = 3 - b$

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

$7 = 6 - b$

$a = 3 - b$

Step 1.3.3.3.3

Simplify the right side.

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Step 1.3.3.3.3.1

Divide $2$ by $- 2$ .

$b = - 1$

$7 = 6 - b$

$a = 3 - b$

$b = - 1$

$7 = 6 - b$

$a = 3 - b$

$b = - 1$

$7 = 6 - b$

$a = 3 - b$

$b = - 1$

$7 = 6 - b$

$a = 3 - b$

Step 1.3.4

Replace all occurrences of $b$ with $- 1$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $b$ in $7 = 6 - b$ with $- 1$ .

$7 = 6 - (- 1)$

$b = - 1$

$a = 3 - b$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $6 - (- 1)$ .

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Step 1.3.4.2.1.1

Multiply $- 1$ by $- 1$ .

$7 = 6 + 1$

$b = - 1$

$a = 3 - b$

Step 1.3.4.2.1.2

Add $6$ and $1$ .

$7 = 7$

$b = - 1$

$a = 3 - b$

$7 = 7$

$b = - 1$

$a = 3 - b$

$7 = 7$

$b = - 1$

$a = 3 - b$

Step 1.3.4.3

Replace all occurrences of $b$ in $a = 3 - b$ with $- 1$ .

$a = 3 - (- 1)$

$7 = 7$

$b = - 1$

Step 1.3.4.4

Simplify the right side.

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Step 1.3.4.4.1

Simplify $3 - (- 1)$ .

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Step 1.3.4.4.1.1

Multiply $- 1$ by $- 1$ .

$a = 3 + 1$

$7 = 7$

$b = - 1$

Step 1.3.4.4.1.2

Add $3$ and $1$ .

$a = 4$

$7 = 7$

$b = - 1$

$a = 4$

$7 = 7$

$b = - 1$

$a = 4$

$7 = 7$

$b = - 1$

$a = 4$

$7 = 7$

$b = - 1$

Step 1.3.5

Remove any equations from the system that are always true.

$a = 4$

$b = - 1$

Step 1.3.6

List all of the solutions.

$a = 4, b = - 1$

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 $a = 4$ , $b = - 1$ , and $x = 1$ .

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Step 1.4.1.1

Multiply $4$ by $1$ .

$y = 4 - 1$

Step 1.4.1.2

Subtract $1$ from $4$ .

$y = 3$

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 $y = 3$ and $y = 3$ .

$3 = 3$

Step 1.4.3

Calculate the value of $y$ when $a = 4$ , $b = - 1$ , and $x = 2$ .

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Step 1.4.3.1

Multiply $4$ by $2$ .

$y = 8 - 1$

Step 1.4.3.2

Subtract $1$ from $8$ .

$y = 7$

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 $y = 7$ and $y = 7$ .

$7 = 7$

Step 1.4.5

Calculate the value of $y$ when $a = 4$ , $b = - 1$ , and $x = 3$ .

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Step 1.4.5.1

Multiply $4$ by $3$ .

$y = 12 - 1$

Step 1.4.5.2

Subtract $1$ from $12$ .

$y = 11$

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 $y = 11$ and $y = 11$ .

$11 = 11$

Step 1.4.7

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 $y = 4 x - 1$ .

$y = 4 x - 1$