Finite Math Examples

$2 y = x + 3$ , $5 y = x - 7$

Step 1

Divide each term in $2 y = x + 3$ by $2$ and simplify.

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Step 1.1

Divide each term in $2 y = x + 3$ by $2$ .

$\frac{2 y}{2} = \frac{x}{2} + \frac{3}{2}$

$5 y = x - 7$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $2$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{2 y}{2} = \frac{x}{2} + \frac{3}{2}$

$5 y = x - 7$

Step 1.2.1.2

Divide $y$ by $1$ .

$y = \frac{x}{2} + \frac{3}{2}$

$5 y = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$5 y = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$5 y = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$5 y = x - 7$

Step 2

Replace all occurrences of $y$ with $\frac{x}{2} + \frac{3}{2}$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $5 y = x - 7$ with $\frac{x}{2} + \frac{3}{2}$ .

$5 (\frac{x}{2} + \frac{3}{2}) = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

Apply the distributive property.

$5 (\frac{x}{2}) + 5 (\frac{3}{2}) = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 2.2.1.2

Combine $5$ and $\frac{x}{2}$ .

$\frac{5 x}{2} + 5 (\frac{3}{2}) = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 2.2.1.3

Multiply $5 (\frac{3}{2})$ .

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Step 2.2.1.3.1

Combine $5$ and $\frac{3}{2}$ .

$\frac{5 x}{2} + \frac{5 \cdot 3}{2} = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 2.2.1.3.2

Multiply $5$ by $3$ .

$\frac{5 x}{2} + \frac{15}{2} = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$\frac{5 x}{2} + \frac{15}{2} = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$\frac{5 x}{2} + \frac{15}{2} = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$\frac{5 x}{2} + \frac{15}{2} = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$\frac{5 x}{2} + \frac{15}{2} = x - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3

Solve for $x$ in $\frac{5 x}{2} + \frac{15}{2} = x - 7$ .

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Step 3.1

Move all terms containing $x$ to the left side of the equation.

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Step 3.1.1

Subtract $x$ from both sides of the equation.

$\frac{5 x}{2} + \frac{15}{2} - x = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.1.2

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

$\frac{5 x}{2} - x \cdot \frac{2}{2} + \frac{15}{2} = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.1.3

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

$\frac{5 x}{2} + \frac{- x \cdot 2}{2} + \frac{15}{2} = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.1.4

Combine the numerators over the common denominator.

$\frac{5 x - x \cdot 2}{2} + \frac{15}{2} = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.1.5

Combine the numerators over the common denominator.

$\frac{5 x - x \cdot 2 + 15}{2} = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.1.6

Multiply $2$ by $- 1$ .

$\frac{5 x - 2 x + 15}{2} = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.1.7

Subtract $2 x$ from $5 x$ .

$\frac{3 x + 15}{2} = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.1.8

Factor $3$ out of $3 x + 15$ .

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Step 3.1.8.1

Factor $3$ out of $3 x$ .

$\frac{3 (x) + 15}{2} = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.1.8.2

Factor $3$ out of $15$ .

$\frac{3 x + 3 \cdot 5}{2} = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.1.8.3

Factor $3$ out of $3 x + 3 \cdot 5$ .

$\frac{3 (x + 5)}{2} = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$\frac{3 (x + 5)}{2} = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$\frac{3 (x + 5)}{2} = - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.2

Multiply both sides of the equation by $\frac{2}{3}$ .

$\frac{2}{3} \cdot \frac{3 (x + 5)}{2} = \frac{2}{3} \cdot - 7$

$y = \frac{x}{2} + \frac{3}{2}$

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

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Step 3.3.1.1.1

Cancel the common factor of $2$ .

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Step 3.3.1.1.1.1

Cancel the common factor.

$\frac{2}{3} \cdot \frac{3 (x + 5)}{2} = \frac{2}{3} \cdot - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.3.1.1.1.2

Rewrite the expression.

$\frac{1}{3} \cdot (3 (x + 5)) = \frac{2}{3} \cdot - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$\frac{1}{3} \cdot (3 (x + 5)) = \frac{2}{3} \cdot - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.3.1.1.2

Cancel the common factor of $3$ .

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Step 3.3.1.1.2.1

Cancel the common factor.

$\frac{1}{3} \cdot (3 (x + 5)) = \frac{2}{3} \cdot - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.3.1.1.2.2

Rewrite the expression.

$x + 5 = \frac{2}{3} \cdot - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$x + 5 = \frac{2}{3} \cdot - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$x + 5 = \frac{2}{3} \cdot - 7$

$y = \frac{x}{2} + \frac{3}{2}$

$x + 5 = \frac{2}{3} \cdot - 7$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.3.2

Simplify the right side.

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Step 3.3.2.1

Simplify $\frac{2}{3} \cdot - 7$ .

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Step 3.3.2.1.1

Multiply $\frac{2}{3} \cdot - 7$ .

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Step 3.3.2.1.1.1

Combine $\frac{2}{3}$ and $- 7$ .

$x + 5 = \frac{2 \cdot - 7}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.3.2.1.1.2

Multiply $2$ by $- 7$ .

$x + 5 = \frac{- 14}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

$x + 5 = \frac{- 14}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.3.2.1.2

Move the negative in front of the fraction.

$x + 5 = - \frac{14}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

$x + 5 = - \frac{14}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

$x + 5 = - \frac{14}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

$x + 5 = - \frac{14}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.4

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

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Step 3.4.1

Subtract $5$ from both sides of the equation.

$x = - \frac{14}{3} - 5$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.4.2

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = - \frac{14}{3} - 5 \cdot \frac{3}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.4.3

Combine $- 5$ and $\frac{3}{3}$ .

$x = - \frac{14}{3} + \frac{- 5 \cdot 3}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.4.4

Combine the numerators over the common denominator.

$x = \frac{- 14 - 5 \cdot 3}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.4.5

Simplify the numerator.

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Step 3.4.5.1

Multiply $- 5$ by $3$ .

$x = \frac{- 14 - 15}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.4.5.2

Subtract $15$ from $- 14$ .

$x = \frac{- 29}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

$x = \frac{- 29}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

Step 3.4.6

Move the negative in front of the fraction.

$x = - \frac{29}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

$x = - \frac{29}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

$x = - \frac{29}{3}$

$y = \frac{x}{2} + \frac{3}{2}$

Step 4

Replace all occurrences of $x$ with $- \frac{29}{3}$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $y = \frac{x}{2} + \frac{3}{2}$ with $- \frac{29}{3}$ .

$y = \frac{- \frac{29}{3}}{2} + \frac{3}{2}$

$x = - \frac{29}{3}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{- \frac{29}{3}}{2} + \frac{3}{2}$ .

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Step 4.2.1.1

Combine the numerators over the common denominator.

$y = \frac{- \frac{29}{3} + 3}{2}$

$x = - \frac{29}{3}$

Step 4.2.1.2

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

$y = \frac{- \frac{29}{3} + 3 \cdot \frac{3}{3}}{2}$

$x = - \frac{29}{3}$

Step 4.2.1.3

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

$y = \frac{- \frac{29}{3} + \frac{3 \cdot 3}{3}}{2}$

$x = - \frac{29}{3}$

Step 4.2.1.4

Combine the numerators over the common denominator.

$y = \frac{\frac{- 29 + 3 \cdot 3}{3}}{2}$

$x = - \frac{29}{3}$

Step 4.2.1.5

Simplify the numerator.

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Step 4.2.1.5.1

Multiply $3$ by $3$ .

$y = \frac{\frac{- 29 + 9}{3}}{2}$

$x = - \frac{29}{3}$

Step 4.2.1.5.2

Add $- 29$ and $9$ .

$y = \frac{\frac{- 20}{3}}{2}$

$x = - \frac{29}{3}$

$y = \frac{\frac{- 20}{3}}{2}$

$x = - \frac{29}{3}$

Step 4.2.1.6

Move the negative in front of the fraction.

$y = \frac{- \frac{20}{3}}{2}$

$x = - \frac{29}{3}$

Step 4.2.1.7

Multiply the numerator by the reciprocal of the denominator.

$y = - \frac{20}{3} \cdot \frac{1}{2}$

$x = - \frac{29}{3}$

Step 4.2.1.8

Cancel the common factor of $2$ .

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Step 4.2.1.8.1

Move the leading negative in $- \frac{20}{3}$ into the numerator.

$y = \frac{- 20}{3} \cdot \frac{1}{2}$

$x = - \frac{29}{3}$

Step 4.2.1.8.2

Factor $2$ out of $- 20$ .

$y = \frac{2 (- 10)}{3} \cdot \frac{1}{2}$

$x = - \frac{29}{3}$

Step 4.2.1.8.3

Cancel the common factor.

$y = \frac{2 \cdot - 10}{3} \cdot \frac{1}{2}$

$x = - \frac{29}{3}$

Step 4.2.1.8.4

Rewrite the expression.

$y = \frac{- 10}{3}$

$x = - \frac{29}{3}$

$y = \frac{- 10}{3}$

$x = - \frac{29}{3}$

Step 4.2.1.9

Move the negative in front of the fraction.

$y = - \frac{10}{3}$

$x = - \frac{29}{3}$

$y = - \frac{10}{3}$

$x = - \frac{29}{3}$

$y = - \frac{10}{3}$

$x = - \frac{29}{3}$

$y = - \frac{10}{3}$

$x = - \frac{29}{3}$

Step 5

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

$(- \frac{29}{3}, - \frac{10}{3})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- \frac{29}{3}, - \frac{10}{3})$

Equation Form:

$x = - \frac{29}{3}, y = - \frac{10}{3}$

Step 7