Finite Math Examples

$\frac{5}{4} x - \frac{2}{3} y = 3$ , $\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1

Solve for $x$ in $\frac{5}{4} x - \frac{2}{3} y = 3$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

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

$\frac{5 x}{4} - \frac{2}{3} y = 3$

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.1.2

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

$\frac{5 x}{4} - \frac{y \cdot 2}{3} = 3$

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.1.3

Move $2$ to the left of $y$ .

$\frac{5 x}{4} - \frac{2 y}{3} = 3$

$\frac{1}{4} x + \frac{5}{3} y = 6$

$\frac{5 x}{4} - \frac{2 y}{3} = 3$

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.2

Add $\frac{2 y}{3}$ to both sides of the equation.

$\frac{5 x}{4} = 3 + \frac{2 y}{3}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.3

Multiply both sides of the equation by $\frac{4}{5}$ .

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

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.4

Simplify both sides of the equation.

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Step 1.4.1

Simplify the left side.

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Step 1.4.1.1

Simplify $\frac{4}{5} \cdot \frac{5 x}{4}$ .

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Step 1.4.1.1.1

Cancel the common factor of $4$ .

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Step 1.4.1.1.1.1

Cancel the common factor.

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

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.4.1.1.1.2

Rewrite the expression.

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

$\frac{1}{4} x + \frac{5}{3} y = 6$

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

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.4.1.1.2

Cancel the common factor of $5$ .

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Step 1.4.1.1.2.1

Factor $5$ out of $5 x$ .

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

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.4.1.1.2.2

Cancel the common factor.

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

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.4.1.1.2.3

Rewrite the expression.

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

$\frac{1}{4} x + \frac{5}{3} y = 6$

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

$\frac{1}{4} x + \frac{5}{3} y = 6$

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

$\frac{1}{4} x + \frac{5}{3} y = 6$

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

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Simplify $\frac{4}{5} (3 + \frac{2 y}{3})$ .

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Step 1.4.2.1.1

Apply the distributive property.

$x = \frac{4}{5} \cdot 3 + \frac{4}{5} \cdot \frac{2 y}{3}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.4.2.1.2

Multiply $\frac{4}{5} \cdot 3$ .

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Step 1.4.2.1.2.1

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

$x = \frac{4 \cdot 3}{5} + \frac{4}{5} \cdot \frac{2 y}{3}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.4.2.1.2.2

Multiply $4$ by $3$ .

$x = \frac{12}{5} + \frac{4}{5} \cdot \frac{2 y}{3}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

$x = \frac{12}{5} + \frac{4}{5} \cdot \frac{2 y}{3}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.4.2.1.3

Multiply $\frac{4}{5} \cdot \frac{2 y}{3}$ .

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Step 1.4.2.1.3.1

Multiply $\frac{4}{5}$ by $\frac{2 y}{3}$ .

$x = \frac{12}{5} + \frac{4 (2 y)}{5 \cdot 3}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.4.2.1.3.2

Multiply $2$ by $4$ .

$x = \frac{12}{5} + \frac{8 y}{5 \cdot 3}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.4.2.1.3.3

Multiply $5$ by $3$ .

$x = \frac{12}{5} + \frac{8 y}{15}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

$x = \frac{12}{5} + \frac{8 y}{15}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

$x = \frac{12}{5} + \frac{8 y}{15}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

$x = \frac{12}{5} + \frac{8 y}{15}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

$x = \frac{12}{5} + \frac{8 y}{15}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 1.5

Reorder $\frac{12}{5}$ and $\frac{8 y}{15}$ .

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{1}{4} x + \frac{5}{3} y = 6$

Step 2

Replace all occurrences of $x$ with $\frac{8 y}{15} + \frac{12}{5}$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $\frac{1}{4} x + \frac{5}{3} y = 6$ with $\frac{8 y}{15} + \frac{12}{5}$ .

$\frac{1}{4} \cdot (\frac{8 y}{15} + \frac{12}{5}) + \frac{5}{3} y = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $\frac{1}{4} \cdot (\frac{8 y}{15} + \frac{12}{5}) + \frac{5}{3} y$ .

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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.

$\frac{1}{4} \cdot \frac{8 y}{15} + \frac{1}{4} \cdot \frac{12}{5} + \frac{5}{3} y = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.1.2

Cancel the common factor of $4$ .

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Step 2.2.1.1.2.1

Factor $4$ out of $8 y$ .

$\frac{1}{4} \cdot \frac{4 (2 y)}{15} + \frac{1}{4} \cdot \frac{12}{5} + \frac{5}{3} y = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.1.2.2

Cancel the common factor.

$\frac{1}{4} \cdot \frac{4 (2 y)}{15} + \frac{1}{4} \cdot \frac{12}{5} + \frac{5}{3} y = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.1.2.3

Rewrite the expression.

$\frac{2 y}{15} + \frac{1}{4} \cdot \frac{12}{5} + \frac{5}{3} y = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{2 y}{15} + \frac{1}{4} \cdot \frac{12}{5} + \frac{5}{3} y = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.1.3

Cancel the common factor of $4$ .

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Step 2.2.1.1.3.1

Factor $4$ out of $12$ .

$\frac{2 y}{15} + \frac{1}{4} \cdot \frac{4 (3)}{5} + \frac{5}{3} y = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.1.3.2

Cancel the common factor.

$\frac{2 y}{15} + \frac{1}{4} \cdot \frac{4 \cdot 3}{5} + \frac{5}{3} y = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.1.3.3

Rewrite the expression.

$\frac{2 y}{15} + \frac{3}{5} + \frac{5}{3} y = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{2 y}{15} + \frac{3}{5} + \frac{5}{3} y = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.1.4

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

$\frac{2 y}{15} + \frac{3}{5} + \frac{5 y}{3} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{2 y}{15} + \frac{3}{5} + \frac{5 y}{3} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.2

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

$\frac{2 y}{15} + \frac{5 y}{3} \cdot \frac{5}{5} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.3

Write each expression with a common denominator of $15$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.2.1.3.1

Multiply $\frac{5 y}{3}$ by $\frac{5}{5}$ .

$\frac{2 y}{15} + \frac{5 y \cdot 5}{3 \cdot 5} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.3.2

Multiply $3$ by $5$ .

$\frac{2 y}{15} + \frac{5 y \cdot 5}{15} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{2 y}{15} + \frac{5 y \cdot 5}{15} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.4

Combine the numerators over the common denominator.

$\frac{2 y + 5 y \cdot 5}{15} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.5

Simplify each term.

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Step 2.2.1.5.1

Simplify the numerator.

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Step 2.2.1.5.1.1

Factor $y$ out of $2 y + 5 y \cdot 5$ .

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Step 2.2.1.5.1.1.1

Factor $y$ out of $2 y$ .

$\frac{y \cdot 2 + 5 y \cdot 5}{15} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.5.1.1.2

Factor $y$ out of $5 y \cdot 5$ .

$\frac{y \cdot 2 + y (5 \cdot 5)}{15} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.5.1.1.3

Factor $y$ out of $y \cdot 2 + y (5 \cdot 5)$ .

$\frac{y (2 + 5 \cdot 5)}{15} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{y (2 + 5 \cdot 5)}{15} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.5.1.2

Multiply $5$ by $5$ .

$\frac{y (2 + 25)}{15} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.5.1.3

Add $2$ and $25$ .

$\frac{y \cdot 27}{15} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{y \cdot 27}{15} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.5.2

Cancel the common factor of $27$ and $15$ .

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Step 2.2.1.5.2.1

Factor $3$ out of $y \cdot 27$ .

$\frac{3 (y \cdot 9)}{15} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.5.2.2

Cancel the common factors.

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Step 2.2.1.5.2.2.1

Factor $3$ out of $15$ .

$\frac{3 (y \cdot 9)}{3 (5)} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.5.2.2.2

Cancel the common factor.

$\frac{3 (y \cdot 9)}{3 \cdot 5} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.5.2.2.3

Rewrite the expression.

$\frac{y \cdot 9}{5} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{y \cdot 9}{5} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{y \cdot 9}{5} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 2.2.1.5.3

Move $9$ to the left of $y$ .

$\frac{9 y}{5} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{9 y}{5} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{9 y}{5} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{9 y}{5} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{9 y}{5} + \frac{3}{5} = 6$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 3

Solve for $y$ in $\frac{9 y}{5} + \frac{3}{5} = 6$ .

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Step 3.1

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

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Step 3.1.1

Subtract $\frac{3}{5}$ from both sides of the equation.

$\frac{9 y}{5} = 6 - \frac{3}{5}$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 3.1.2

To write $6$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{9 y}{5} = 6 \cdot \frac{5}{5} - \frac{3}{5}$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 3.1.3

Combine $6$ and $\frac{5}{5}$ .

$\frac{9 y}{5} = \frac{6 \cdot 5}{5} - \frac{3}{5}$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 3.1.4

Combine the numerators over the common denominator.

$\frac{9 y}{5} = \frac{6 \cdot 5 - 3}{5}$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 3.1.5

Simplify the numerator.

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Step 3.1.5.1

Multiply $6$ by $5$ .

$\frac{9 y}{5} = \frac{30 - 3}{5}$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 3.1.5.2

Subtract $3$ from $30$ .

$\frac{9 y}{5} = \frac{27}{5}$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{9 y}{5} = \frac{27}{5}$

$x = \frac{8 y}{15} + \frac{12}{5}$

$\frac{9 y}{5} = \frac{27}{5}$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 3.2

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$9 y = 27$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 3.3

Divide each term in $9 y = 27$ by $9$ and simplify.

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Step 3.3.1

Divide each term in $9 y = 27$ by $9$ .

$\frac{9 y}{9} = \frac{27}{9}$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $9$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{9 y}{9} = \frac{27}{9}$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 3.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{27}{9}$

$x = \frac{8 y}{15} + \frac{12}{5}$

$y = \frac{27}{9}$

$x = \frac{8 y}{15} + \frac{12}{5}$

$y = \frac{27}{9}$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide $27$ by $9$ .

$y = 3$

$x = \frac{8 y}{15} + \frac{12}{5}$

$y = 3$

$x = \frac{8 y}{15} + \frac{12}{5}$

$y = 3$

$x = \frac{8 y}{15} + \frac{12}{5}$

$y = 3$

$x = \frac{8 y}{15} + \frac{12}{5}$

Step 4

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

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Step 4.1

Replace all occurrences of $y$ in $x = \frac{8 y}{15} + \frac{12}{5}$ with $3$ .

$x = \frac{8 (3)}{15} + \frac{12}{5}$

$y = 3$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{8 (3)}{15} + \frac{12}{5}$ .

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Step 4.2.1.1

Multiply $8$ by $3$ .

$x = \frac{24}{15} + \frac{12}{5}$

$y = 3$

Step 4.2.1.2

Cancel the common factor of $24$ and $15$ .

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Step 4.2.1.2.1

Factor $3$ out of $24$ .

$x = \frac{3 (8)}{15} + \frac{12}{5}$

$y = 3$

Step 4.2.1.2.2

Cancel the common factors.

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Step 4.2.1.2.2.1

Factor $3$ out of $15$ .

$x = \frac{3 \cdot 8}{3 \cdot 5} + \frac{12}{5}$

$y = 3$

Step 4.2.1.2.2.2

Cancel the common factor.

$x = \frac{3 \cdot 8}{3 \cdot 5} + \frac{12}{5}$

$y = 3$

Step 4.2.1.2.2.3

Rewrite the expression.

$x = \frac{8}{5} + \frac{12}{5}$

$y = 3$

$x = \frac{8}{5} + \frac{12}{5}$

$y = 3$

$x = \frac{8}{5} + \frac{12}{5}$

$y = 3$

Step 4.2.1.3

Combine the numerators over the common denominator.

$x = \frac{8 + 12}{5}$

$y = 3$

Step 4.2.1.4

Simplify the expression.

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Step 4.2.1.4.1

Add $8$ and $12$ .

$x = \frac{20}{5}$

$y = 3$

Step 4.2.1.4.2

Divide $20$ by $5$ .

$x = 4$

$y = 3$

$x = 4$

$y = 3$

$x = 4$

$y = 3$

$x = 4$

$y = 3$

$x = 4$

$y = 3$

Step 5

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

$(4, 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 3)$

Equation Form:

$x = 4, y = 3$

Step 7