Finite Math Examples

$3 x - 8 y = 5$ , $4 x - 4 y = 5$

Step 1

Solve for $x$ in $3 x - 8 y = 5$ .

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Step 1.1

Add $8 y$ to both sides of the equation.

$3 x = 5 + 8 y$

$4 x - 4 y = 5$

Step 1.2

Divide each term in $3 x = 5 + 8 y$ by $3$ and simplify.

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Step 1.2.1

Divide each term in $3 x = 5 + 8 y$ by $3$ .

$\frac{3 x}{3} = \frac{5}{3} + \frac{8 y}{3}$

$4 x - 4 y = 5$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of $3$ .

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Step 1.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{5}{3} + \frac{8 y}{3}$

$4 x - 4 y = 5$

Step 1.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{3} + \frac{8 y}{3}$

$4 x - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$4 x - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$4 x - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$4 x - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$4 x - 4 y = 5$

Step 2

Replace all occurrences of $x$ with $\frac{5}{3} + \frac{8 y}{3}$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $4 x - 4 y = 5$ with $\frac{5}{3} + \frac{8 y}{3}$ .

$4 (\frac{5}{3} + \frac{8 y}{3}) - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $4 (\frac{5}{3} + \frac{8 y}{3}) - 4 y$ .

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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.

$4 (\frac{5}{3}) + 4 (\frac{8 y}{3}) - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.1.2

Multiply $4 (\frac{5}{3})$ .

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Step 2.2.1.1.2.1

Combine $4$ and $\frac{5}{3}$ .

$\frac{4 \cdot 5}{3} + 4 (\frac{8 y}{3}) - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.1.2.2

Multiply $4$ by $5$ .

$\frac{20}{3} + 4 (\frac{8 y}{3}) - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$\frac{20}{3} + 4 (\frac{8 y}{3}) - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.1.3

Multiply $4 \frac{8 y}{3}$ .

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Step 2.2.1.1.3.1

Combine $4$ and $\frac{8 y}{3}$ .

$\frac{20}{3} + \frac{4 (8 y)}{3} - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.1.3.2

Multiply $8$ by $4$ .

$\frac{20}{3} + \frac{32 y}{3} - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$\frac{20}{3} + \frac{32 y}{3} - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$\frac{20}{3} + \frac{32 y}{3} - 4 y = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.2

To write $- 4 y$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{20}{3} + \frac{32 y}{3} - 4 y \cdot \frac{3}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.3

Combine $- 4 y$ and $\frac{3}{3}$ .

$\frac{20}{3} + \frac{32 y}{3} + \frac{- 4 y \cdot 3}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.4

Combine the numerators over the common denominator.

$\frac{20}{3} + \frac{32 y - 4 y \cdot 3}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.5

Combine the numerators over the common denominator.

$\frac{20 + 32 y - 4 y \cdot 3}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.6

Multiply $3$ by $- 4$ .

$\frac{20 + 32 y - 12 y}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.7

Subtract $12 y$ from $32 y$ .

$\frac{20 + 20 y}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.8

Factor $20$ out of $20 + 20 y$ .

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Step 2.2.1.8.1

Factor $20$ out of $20$ .

$\frac{20 \cdot 1 + 20 y}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 2.2.1.8.2

Factor $20$ out of $20 \cdot 1 + 20 y$ .

$\frac{20 (1 + y)}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$\frac{20 (1 + y)}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$\frac{20 (1 + y)}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$\frac{20 (1 + y)}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$\frac{20 (1 + y)}{3} = 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3

Solve for $y$ in $\frac{20 (1 + y)}{3} = 5$ .

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Step 3.1

Multiply both sides of the equation by $\frac{3}{20}$ .

$\frac{3}{20} \cdot \frac{20 (1 + y)}{3} = \frac{3}{20} \cdot 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.2

Simplify both sides of the equation.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Simplify $\frac{3}{20} \cdot \frac{20 (1 + y)}{3}$ .

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Step 3.2.1.1.1

Cancel the common factor of $3$ .

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Step 3.2.1.1.1.1

Cancel the common factor.

$\frac{3}{20} \cdot \frac{20 (1 + y)}{3} = \frac{3}{20} \cdot 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.2.1.1.1.2

Rewrite the expression.

$\frac{1}{20} \cdot (20 (1 + y)) = \frac{3}{20} \cdot 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$\frac{1}{20} \cdot (20 (1 + y)) = \frac{3}{20} \cdot 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.2.1.1.2

Cancel the common factor of $20$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

$\frac{1}{20} \cdot (20 (1 + y)) = \frac{3}{20} \cdot 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.2.1.1.2.2

Rewrite the expression.

$1 + y = \frac{3}{20} \cdot 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$1 + y = \frac{3}{20} \cdot 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$1 + y = \frac{3}{20} \cdot 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

$1 + y = \frac{3}{20} \cdot 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Cancel the common factor of $5$ .

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Step 3.2.2.1.1

Factor $5$ out of $20$ .

$1 + y = \frac{3}{5 (4)} \cdot 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.2.2.1.2

Cancel the common factor.

$1 + y = \frac{3}{5 \cdot 4} \cdot 5$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.2.2.1.3

Rewrite the expression.

$1 + y = \frac{3}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

$1 + y = \frac{3}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

$1 + y = \frac{3}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

$1 + y = \frac{3}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.3

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

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Step 3.3.1

Subtract $1$ from both sides of the equation.

$y = \frac{3}{4} - 1$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.3.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$y = \frac{3}{4} - 1 \cdot \frac{4}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.3.3

Combine $- 1$ and $\frac{4}{4}$ .

$y = \frac{3}{4} + \frac{- 1 \cdot 4}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.3.4

Combine the numerators over the common denominator.

$y = \frac{3 - 1 \cdot 4}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.3.5

Simplify the numerator.

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Step 3.3.5.1

Multiply $- 1$ by $4$ .

$y = \frac{3 - 4}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.3.5.2

Subtract $4$ from $3$ .

$y = \frac{- 1}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

$y = \frac{- 1}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 3.3.6

Move the negative in front of the fraction.

$y = - \frac{1}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

$y = - \frac{1}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

$y = - \frac{1}{4}$

$x = \frac{5}{3} + \frac{8 y}{3}$

Step 4

Replace all occurrences of $y$ with $- \frac{1}{4}$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = \frac{5}{3} + \frac{8 y}{3}$ with $- \frac{1}{4}$ .

$x = \frac{5}{3} + \frac{8 (- \frac{1}{4})}{3}$

$y = - \frac{1}{4}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{5}{3} + \frac{8 (- \frac{1}{4})}{3}$ .

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Step 4.2.1.1

Combine the numerators over the common denominator.

$x = \frac{5 + 8 (- \frac{1}{4})}{3}$

$y = - \frac{1}{4}$

Step 4.2.1.2

Simplify each term.

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Step 4.2.1.2.1

Cancel the common factor of $4$ .

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Step 4.2.1.2.1.1

Move the leading negative in $- \frac{1}{4}$ into the numerator.

$x = \frac{5 + 8 (\frac{- 1}{4})}{3}$

$y = - \frac{1}{4}$

Step 4.2.1.2.1.2

Factor $4$ out of $8$ .

$x = \frac{5 + 4 (2) (\frac{- 1}{4})}{3}$

$y = - \frac{1}{4}$

Step 4.2.1.2.1.3

Cancel the common factor.

$x = \frac{5 + 4 \cdot (2 (\frac{- 1}{4}))}{3}$

$y = - \frac{1}{4}$

Step 4.2.1.2.1.4

Rewrite the expression.

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

$y = - \frac{1}{4}$

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

$y = - \frac{1}{4}$

Step 4.2.1.2.2

Multiply $2$ by $- 1$ .

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

$y = - \frac{1}{4}$

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

$y = - \frac{1}{4}$

Step 4.2.1.3

Simplify the expression.

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Step 4.2.1.3.1

Subtract $2$ from $5$ .

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

$y = - \frac{1}{4}$

Step 4.2.1.3.2

Divide $3$ by $3$ .

$x = 1$

$y = - \frac{1}{4}$

$x = 1$

$y = - \frac{1}{4}$

$x = 1$

$y = - \frac{1}{4}$

$x = 1$

$y = - \frac{1}{4}$

$x = 1$

$y = - \frac{1}{4}$

Step 5

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

$(1, - \frac{1}{4})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(1, - \frac{1}{4})$

Equation Form:

$x = 1, y = - \frac{1}{4}$

Step 7