Finite Math Examples

$y = - \frac{1}{3} x + 5$ , $2 x - 2 y = - 2$

Step 1

Combine $x$ and $\frac{1}{3}$ .

$y = - \frac{x}{3} + 5$

$2 x - 2 y = - 2$

Step 2

Replace all occurrences of $y$ with $- \frac{x}{3} + 5$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $2 x - 2 y = - 2$ with $- \frac{x}{3} + 5$ .

$2 x - 2 (- \frac{x}{3} + 5) = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $2 x - 2 (- \frac{x}{3} + 5)$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Apply the distributive property.

$2 x - 2 (- \frac{x}{3}) - 2 \cdot 5 = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2.1.1.2

Multiply $- 2 (- \frac{x}{3})$ .

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Step 2.2.1.1.2.1

Multiply $- 1$ by $- 2$ .

$2 x + 2 (\frac{x}{3}) - 2 \cdot 5 = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2.1.1.2.2

Combine $2$ and $\frac{x}{3}$ .

$2 x + \frac{2 x}{3} - 2 \cdot 5 = - 2$

$y = - \frac{x}{3} + 5$

$2 x + \frac{2 x}{3} - 2 \cdot 5 = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2.1.1.3

Multiply $- 2$ by $5$ .

$2 x + \frac{2 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

$2 x + \frac{2 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2.1.2

To write $2 x$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$2 x \cdot \frac{3}{3} + \frac{2 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2.1.3

Simplify terms.

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Step 2.2.1.3.1

Combine $2 x$ and $\frac{3}{3}$ .

$\frac{2 x \cdot 3}{3} + \frac{2 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2.1.3.2

Combine the numerators over the common denominator.

$\frac{2 x \cdot 3 + 2 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

$\frac{2 x \cdot 3 + 2 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2.1.4

Simplify the numerator.

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Step 2.2.1.4.1

Factor $2 x$ out of $2 x \cdot 3 + 2 x$ .

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Step 2.2.1.4.1.1

Factor $2 x$ out of $2 x \cdot 3$ .

$\frac{2 x (3) + 2 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2.1.4.1.2

Factor $2 x$ out of $2 x$ .

$\frac{2 x (3) + 2 x (1)}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2.1.4.1.3

Factor $2 x$ out of $2 x (3) + 2 x (1)$ .

$\frac{2 x (3 + 1)}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

$\frac{2 x (3 + 1)}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2.1.4.2

Add $3$ and $1$ .

$\frac{2 x \cdot 4}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

Step 2.2.1.4.3

Multiply $4$ by $2$ .

$\frac{8 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

$\frac{8 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

$\frac{8 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

$\frac{8 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

$\frac{8 x}{3} - 10 = - 2$

$y = - \frac{x}{3} + 5$

Step 3

Solve for $x$ in $\frac{8 x}{3} - 10 = - 2$ .

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Step 3.1

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

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Step 3.1.1

Add $10$ to both sides of the equation.

$\frac{8 x}{3} = - 2 + 10$

$y = - \frac{x}{3} + 5$

Step 3.1.2

Add $- 2$ and $10$ .

$\frac{8 x}{3} = 8$

$y = - \frac{x}{3} + 5$

$\frac{8 x}{3} = 8$

$y = - \frac{x}{3} + 5$

Step 3.2

Multiply both sides of the equation by $\frac{3}{8}$ .

$\frac{3}{8} \cdot \frac{8 x}{3} = \frac{3}{8} \cdot 8$

$y = - \frac{x}{3} + 5$

Step 3.3

Simplify both sides of the equation.

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Step 3.3.1

Simplify the left side.

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Step 3.3.1.1

Simplify $\frac{3}{8} \cdot \frac{8 x}{3}$ .

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Step 3.3.1.1.1

Cancel the common factor of $3$ .

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Step 3.3.1.1.1.1

Cancel the common factor.

$\frac{3}{8} \cdot \frac{8 x}{3} = \frac{3}{8} \cdot 8$

$y = - \frac{x}{3} + 5$

Step 3.3.1.1.1.2

Rewrite the expression.

$\frac{1}{8} \cdot (8 x) = \frac{3}{8} \cdot 8$

$y = - \frac{x}{3} + 5$

$\frac{1}{8} \cdot (8 x) = \frac{3}{8} \cdot 8$

$y = - \frac{x}{3} + 5$

Step 3.3.1.1.2

Cancel the common factor of $8$ .

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Step 3.3.1.1.2.1

Factor $8$ out of $8 x$ .

$\frac{1}{8} \cdot (8 (x)) = \frac{3}{8} \cdot 8$

$y = - \frac{x}{3} + 5$

Step 3.3.1.1.2.2

Cancel the common factor.

$\frac{1}{8} \cdot (8 x) = \frac{3}{8} \cdot 8$

$y = - \frac{x}{3} + 5$

Step 3.3.1.1.2.3

Rewrite the expression.

$x = \frac{3}{8} \cdot 8$

$y = - \frac{x}{3} + 5$

$x = \frac{3}{8} \cdot 8$

$y = - \frac{x}{3} + 5$

$x = \frac{3}{8} \cdot 8$

$y = - \frac{x}{3} + 5$

$x = \frac{3}{8} \cdot 8$

$y = - \frac{x}{3} + 5$

Step 3.3.2

Simplify the right side.

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Step 3.3.2.1

Cancel the common factor of $8$ .

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Step 3.3.2.1.1

Cancel the common factor.

$x = \frac{3}{8} \cdot 8$

$y = - \frac{x}{3} + 5$

Step 3.3.2.1.2

Rewrite the expression.

$x = 3$

$y = - \frac{x}{3} + 5$

$x = 3$

$y = - \frac{x}{3} + 5$

$x = 3$

$y = - \frac{x}{3} + 5$

$x = 3$

$y = - \frac{x}{3} + 5$

$x = 3$

$y = - \frac{x}{3} + 5$

Step 4

Replace all occurrences of $x$ with $3$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $y = - \frac{x}{3} + 5$ with $3$ .

$y = - \frac{3}{3} + 5$

$x = 3$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $- \frac{3}{3} + 5$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Cancel the common factor of $3$ .

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Step 4.2.1.1.1.1

Cancel the common factor.

$y = - \frac{3}{3} + 5$

$x = 3$

Step 4.2.1.1.1.2

Rewrite the expression.

$y = - 1 \cdot 1 + 5$

$x = 3$

$y = - 1 \cdot 1 + 5$

$x = 3$

Step 4.2.1.1.2

Multiply $- 1$ by $1$ .

$y = - 1 + 5$

$x = 3$

$y = - 1 + 5$

$x = 3$

Step 4.2.1.2

Add $- 1$ and $5$ .

$y = 4$

$x = 3$

$y = 4$

$x = 3$

$y = 4$

$x = 3$

$y = 4$

$x = 3$

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(3, 4)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, 4)$

Equation Form:

$x = 3, y = 4$

Step 7