Solve by Addition/Elimination 2/3x+3/5y=12 , 5/2y-3x=6

$\frac{2}{3} x + \frac{3}{5} y = 12$ , $\frac{5}{2} y - 3 x = 6$

Step 1

Multiply each term in $\frac{2}{3} x + \frac{3}{5} y = 12$ by $15$ to eliminate the fractions.

Multiply each term in $\frac{2}{3} x + \frac{3}{5} y = 12$ by $15$ .

$\frac{2}{3} x \cdot 15 + \frac{3}{5} y \cdot 15 = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

Simplify the left side.

Simplify each term.

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

$\frac{2 x}{3} \cdot 15 + \frac{3}{5} y \cdot 15 = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

Cancel the common factor of $3$ .

Factor $3$ out of $15$ .

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

$\frac{5}{2} y - 3 x = 6$

Cancel the common factor.

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

$\frac{5}{2} y - 3 x = 6$

Rewrite the expression.

$2 x \cdot 5 + \frac{3}{5} y \cdot 15 = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

$2 x \cdot 5 + \frac{3}{5} y \cdot 15 = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

Multiply $5$ by $2$ .

$10 x + \frac{3}{5} y \cdot 15 = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

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

$10 x + \frac{3 y}{5} \cdot 15 = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

Cancel the common factor of $5$ .

Factor $5$ out of $15$ .

$10 x + \frac{3 y}{5} \cdot (5 (3)) = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

Cancel the common factor.

$10 x + \frac{3 y}{5} \cdot (5 \cdot 3) = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

Rewrite the expression.

$10 x + 3 y \cdot 3 = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

$10 x + 3 y \cdot 3 = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

Multiply $3$ by $3$ .

$10 x + 9 y = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

$10 x + 9 y = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

$10 x + 9 y = 12 \cdot 15$

$\frac{5}{2} y - 3 x = 6$

Simplify the right side.

Multiply $12$ by $15$ .

$10 x + 9 y = 180$

$\frac{5}{2} y - 3 x = 6$

$10 x + 9 y = 180$

$\frac{5}{2} y - 3 x = 6$

$10 x + 9 y = 180$

$\frac{5}{2} y - 3 x = 6$

Step 2

Multiply each term in $\frac{5}{2} y - 3 x = 6$ by $2$ to eliminate the fractions.

Multiply each term in $\frac{5}{2} y - 3 x = 6$ by $2$ .

$\frac{5}{2} y \cdot 2 - 3 x \cdot 2 = 6 \cdot 2$

$10 x + 9 y = 180$

Simplify the left side.

Simplify each term.

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

$\frac{5 y}{2} \cdot 2 - 3 x \cdot 2 = 6 \cdot 2$

$10 x + 9 y = 180$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{5 y}{2} \cdot 2 - 3 x \cdot 2 = 6 \cdot 2$

$10 x + 9 y = 180$

Rewrite the expression.

$5 y - 3 x \cdot 2 = 6 \cdot 2$

$10 x + 9 y = 180$

$5 y - 3 x \cdot 2 = 6 \cdot 2$

$10 x + 9 y = 180$

Multiply $2$ by $- 3$ .

$5 y - 6 x = 6 \cdot 2$

$10 x + 9 y = 180$

$5 y - 6 x = 6 \cdot 2$

$10 x + 9 y = 180$

$5 y - 6 x = 6 \cdot 2$

$10 x + 9 y = 180$

Simplify the right side.

Multiply $6$ by $2$ .

$5 y - 6 x = 12$

$10 x + 9 y = 180$

$5 y - 6 x = 12$

$10 x + 9 y = 180$

$5 y - 6 x = 12$

$10 x + 9 y = 180$

Step 3

Simplify the left side.

Reorder $5 y$ and $- 6 x$ .

$- 6 x + 5 y = 12$

$10 x + 9 y = 180$

$- 6 x + 5 y = 12$

$10 x + 9 y = 180$

Step 4

Multiply each equation by the value that makes the coefficients of $x$ opposite.

$(3) \cdot (10 x + 9 y) = (3) (180)$

$(5) \cdot (- 6 x + 5 y) = (5) (12)$

Step 5

Simplify.

Simplify the left side.

Simplify $(3) \cdot (10 x + 9 y)$ .

Apply the distributive property.

$3 (10 x) + 3 (9 y) = (3) (180)$

$(5) \cdot (- 6 x + 5 y) = (5) (12)$

Multiply.

Multiply $10$ by $3$ .

$30 x + 3 (9 y) = (3) (180)$

$(5) \cdot (- 6 x + 5 y) = (5) (12)$

Multiply $9$ by $3$ .

$30 x + 27 y = (3) (180)$

$(5) \cdot (- 6 x + 5 y) = (5) (12)$

$30 x + 27 y = (3) (180)$

$(5) \cdot (- 6 x + 5 y) = (5) (12)$

$30 x + 27 y = (3) (180)$

$(5) \cdot (- 6 x + 5 y) = (5) (12)$

$30 x + 27 y = (3) (180)$

$(5) \cdot (- 6 x + 5 y) = (5) (12)$

Simplify the right side.

Multiply $3$ by $180$ .

$30 x + 27 y = 540$

$(5) \cdot (- 6 x + 5 y) = (5) (12)$

$30 x + 27 y = 540$

$(5) \cdot (- 6 x + 5 y) = (5) (12)$

Simplify the left side.

Simplify $(5) \cdot (- 6 x + 5 y)$ .

Apply the distributive property.

$30 x + 27 y = 540$

$5 (- 6 x) + 5 (5 y) = (5) (12)$

Multiply.

Multiply $- 6$ by $5$ .

$30 x + 27 y = 540$

$- 30 x + 5 (5 y) = (5) (12)$

Multiply $5$ by $5$ .

$30 x + 27 y = 540$

$- 30 x + 25 y = (5) (12)$

$30 x + 27 y = 540$

$- 30 x + 25 y = (5) (12)$

$30 x + 27 y = 540$

$- 30 x + 25 y = (5) (12)$

$30 x + 27 y = 540$

$- 30 x + 25 y = (5) (12)$

Simplify the right side.

Multiply $5$ by $12$ .

$30 x + 27 y = 540$

$- 30 x + 25 y = 60$

$30 x + 27 y = 540$

$- 30 x + 25 y = 60$

$30 x + 27 y = 540$

$- 30 x + 25 y = 60$

Step 6

Add the two equations together to eliminate $x$ from the system.

		$3$	$0$	$x$	$+$	$2$	$7$	$y$	$=$	$5$	$4$	$0$
$+$	$-$	$3$	$0$	$x$	$+$	$2$	$5$	$y$	$=$		$6$	$0$
						$5$	$2$	$y$	$=$	$6$	$0$	$0$

Step 7

Divide each term in $52 y = 600$ by $52$ and simplify.

Divide each term in $52 y = 600$ by $52$ .

$\frac{52 y}{52} = \frac{600}{52}$

Simplify the left side.

Cancel the common factor of $52$ .

Cancel the common factor.

$\frac{52 y}{52} = \frac{600}{52}$

Divide $y$ by $1$ .

$y = \frac{600}{52}$

Simplify the right side.

Cancel the common factor of $600$ and $52$ .

Factor $4$ out of $600$ .

$y = \frac{4 (150)}{52}$

Cancel the common factors.

Factor $4$ out of $52$ .

$y = \frac{4 \cdot 150}{4 \cdot 13}$

Cancel the common factor.

$y = \frac{4 \cdot 150}{4 \cdot 13}$

Rewrite the expression.

$y = \frac{150}{13}$

Step 8

Substitute the value found for $y$ into one of the original equations, then solve for $x$ .

Substitute the value found for $y$ into one of the original equations to solve for $x$ .

$30 x + 27 (\frac{150}{13}) = 540$

Multiply $27 (\frac{150}{13})$ .

Combine $27$ and $\frac{150}{13}$ .

$30 x + \frac{27 \cdot 150}{13} = 540$

Multiply $27$ by $150$ .

$30 x + \frac{4050}{13} = 540$

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

Subtract $\frac{4050}{13}$ from both sides of the equation.

$30 x = 540 - \frac{4050}{13}$

To write $540$ as a fraction with a common denominator, multiply by $\frac{13}{13}$ .

$30 x = 540 \cdot \frac{13}{13} - \frac{4050}{13}$

Combine $540$ and $\frac{13}{13}$ .

$30 x = \frac{540 \cdot 13}{13} - \frac{4050}{13}$

Combine the numerators over the common denominator.

$30 x = \frac{540 \cdot 13 - 4050}{13}$

Simplify the numerator.

Multiply $540$ by $13$ .

$30 x = \frac{7020 - 4050}{13}$

Subtract $4050$ from $7020$ .

$30 x = \frac{2970}{13}$

Divide each term in $30 x = \frac{2970}{13}$ by $30$ and simplify.

Divide each term in $30 x = \frac{2970}{13}$ by $30$ .

$\frac{30 x}{30} = \frac{\frac{2970}{13}}{30}$

Simplify the left side.

Cancel the common factor of $30$ .

Cancel the common factor.

$\frac{30 x}{30} = \frac{\frac{2970}{13}}{30}$

Divide $x$ by $1$ .

$x = \frac{\frac{2970}{13}}{30}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{2970}{13} \cdot \frac{1}{30}$

Cancel the common factor of $30$ .

Factor $30$ out of $2970$ .

$x = \frac{30 (99)}{13} \cdot \frac{1}{30}$

Cancel the common factor.

$x = \frac{30 \cdot 99}{13} \cdot \frac{1}{30}$

Rewrite the expression.

$x = \frac{99}{13}$

Step 9

The solution to the independent system of equations can be represented as a point.

$(\frac{99}{13}, \frac{150}{13})$

Step 10

The result can be shown in multiple forms.

Point Form:

$(\frac{99}{13}, \frac{150}{13})$

Equation Form:

$x = \frac{99}{13}, y = \frac{150}{13}$

Step 11

Solve by Addition/Elimination 2/3x+3/5y=12 , 5/2y-3x=6

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