Solve by Addition/Elimination 3x-2y=2 5x-5y=10

$3 x - 2 y = 2$ $5 x - 5 y = 10$

Step 1

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

$(- 5) \cdot (3 x - 2 y) = (- 5) (2)$

$(2) \cdot (5 x - 5 y) = (2) (10)$

Step 2

Simplify.

Simplify the left side.

Simplify $(- 5) \cdot (3 x - 2 y)$ .

Apply the distributive property.

$- 5 (3 x) - 5 (- 2 y) = (- 5) (2)$

$(2) \cdot (5 x - 5 y) = (2) (10)$

Multiply.

Multiply $3$ by $- 5$ .

$- 15 x - 5 (- 2 y) = (- 5) (2)$

$(2) \cdot (5 x - 5 y) = (2) (10)$

Multiply $- 2$ by $- 5$ .

$- 15 x + 10 y = (- 5) (2)$

$(2) \cdot (5 x - 5 y) = (2) (10)$

$- 15 x + 10 y = (- 5) (2)$

$(2) \cdot (5 x - 5 y) = (2) (10)$

$- 15 x + 10 y = (- 5) (2)$

$(2) \cdot (5 x - 5 y) = (2) (10)$

$- 15 x + 10 y = (- 5) (2)$

$(2) \cdot (5 x - 5 y) = (2) (10)$

Simplify the right side.

Multiply $- 5$ by $2$ .

$- 15 x + 10 y = - 10$

$(2) \cdot (5 x - 5 y) = (2) (10)$

$- 15 x + 10 y = - 10$

$(2) \cdot (5 x - 5 y) = (2) (10)$

Simplify the left side.

Simplify $(2) \cdot (5 x - 5 y)$ .

Apply the distributive property.

$- 15 x + 10 y = - 10$

$2 (5 x) + 2 (- 5 y) = (2) (10)$

Multiply.

Multiply $5$ by $2$ .

$- 15 x + 10 y = - 10$

$10 x + 2 (- 5 y) = (2) (10)$

Multiply $- 5$ by $2$ .

$- 15 x + 10 y = - 10$

$10 x - 10 y = (2) (10)$

$- 15 x + 10 y = - 10$

$10 x - 10 y = (2) (10)$

$- 15 x + 10 y = - 10$

$10 x - 10 y = (2) (10)$

$- 15 x + 10 y = - 10$

$10 x - 10 y = (2) (10)$

Simplify the right side.

Multiply $2$ by $10$ .

$- 15 x + 10 y = - 10$

$10 x - 10 y = 20$

$- 15 x + 10 y = - 10$

$10 x - 10 y = 20$

$- 15 x + 10 y = - 10$

$10 x - 10 y = 20$

Step 3

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

	$-$	$1$	$5$	$x$	$+$	$1$	$0$	$y$	$=$	$-$	$1$	$0$
$+$		$1$	$0$	$x$	$-$	$1$	$0$	$y$	$=$		$2$	$0$
		$-$	$5$	$x$					$=$		$1$	$0$

Step 4

Divide each term in $- 5 x = 10$ by $- 5$ and simplify.

Divide each term in $- 5 x = 10$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{10}{- 5}$

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{10}{- 5}$

Divide $x$ by $1$ .

$x = \frac{10}{- 5}$

Simplify the right side.

Divide $10$ by $- 5$ .

$x = - 2$

Step 5

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

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

$- 15 \cdot - 2 + 10 y = - 10$

Multiply $- 15$ by $- 2$ .

$30 + 10 y = - 10$

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

Subtract $30$ from both sides of the equation.

$10 y = - 10 - 30$

Subtract $30$ from $- 10$ .

$10 y = - 40$

Divide each term in $10 y = - 40$ by $10$ and simplify.

Divide each term in $10 y = - 40$ by $10$ .

$\frac{10 y}{10} = \frac{- 40}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 y}{10} = \frac{- 40}{10}$

Divide $y$ by $1$ .

$y = \frac{- 40}{10}$

Simplify the right side.

Divide $- 40$ by $10$ .

$y = - 4$

Step 6

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

$(- 2, - 4)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 2, - 4)$

Equation Form:

$x = - 2, y = - 4$

Step 8

Solve by Addition/Elimination 3x-2y=2 5x-5y=10

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