Solve by Addition/Elimination 3x+8y=15 , 2x-8y=10

$3 x + 8 y = 15$ , $2 x - 8 y = 10$

Step 1

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

$(- 2) \cdot (3 x + 8 y) = (- 2) (15)$

$(3) \cdot (2 x - 8 y) = (3) (10)$

Step 2

Simplify.

Simplify the left side.

Simplify $(- 2) \cdot (3 x + 8 y)$ .

Apply the distributive property.

$- 2 (3 x) - 2 (8 y) = (- 2) (15)$

$(3) \cdot (2 x - 8 y) = (3) (10)$

Multiply.

Multiply $3$ by $- 2$ .

$- 6 x - 2 (8 y) = (- 2) (15)$

$(3) \cdot (2 x - 8 y) = (3) (10)$

Multiply $8$ by $- 2$ .

$- 6 x - 16 y = (- 2) (15)$

$(3) \cdot (2 x - 8 y) = (3) (10)$

$- 6 x - 16 y = (- 2) (15)$

$(3) \cdot (2 x - 8 y) = (3) (10)$

$- 6 x - 16 y = (- 2) (15)$

$(3) \cdot (2 x - 8 y) = (3) (10)$

$- 6 x - 16 y = (- 2) (15)$

$(3) \cdot (2 x - 8 y) = (3) (10)$

Simplify the right side.

Multiply $- 2$ by $15$ .

$- 6 x - 16 y = - 30$

$(3) \cdot (2 x - 8 y) = (3) (10)$

$- 6 x - 16 y = - 30$

$(3) \cdot (2 x - 8 y) = (3) (10)$

Simplify the left side.

Simplify $(3) \cdot (2 x - 8 y)$ .

Apply the distributive property.

$- 6 x - 16 y = - 30$

$3 (2 x) + 3 (- 8 y) = (3) (10)$

Multiply.

Multiply $2$ by $3$ .

$- 6 x - 16 y = - 30$

$6 x + 3 (- 8 y) = (3) (10)$

Multiply $- 8$ by $3$ .

$- 6 x - 16 y = - 30$

$6 x - 24 y = (3) (10)$

$- 6 x - 16 y = - 30$

$6 x - 24 y = (3) (10)$

$- 6 x - 16 y = - 30$

$6 x - 24 y = (3) (10)$

$- 6 x - 16 y = - 30$

$6 x - 24 y = (3) (10)$

Simplify the right side.

Multiply $3$ by $10$ .

$- 6 x - 16 y = - 30$

$6 x - 24 y = 30$

$- 6 x - 16 y = - 30$

$6 x - 24 y = 30$

$- 6 x - 16 y = - 30$

$6 x - 24 y = 30$

Step 3

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

	$-$	$6$	$x$	$-$	$1$	$6$	$y$	$=$	$-$	$3$	$0$
$+$		$6$	$x$	$-$	$2$	$4$	$y$	$=$		$3$	$0$
				$-$	$4$	$0$	$y$	$=$			$0$

Step 4

Divide each term in $- 40 y = 0$ by $- 40$ and simplify.

Divide each term in $- 40 y = 0$ by $- 40$ .

$\frac{- 40 y}{- 40} = \frac{0}{- 40}$

Simplify the left side.

Cancel the common factor of $- 40$ .

Cancel the common factor.

$\frac{- 40 y}{- 40} = \frac{0}{- 40}$

Divide $y$ by $1$ .

$y = \frac{0}{- 40}$

Simplify the right side.

Divide $0$ by $- 40$ .

$y = 0$

Step 5

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 $x$ .

$- 6 x - 16 \cdot 0 = - 30$

Simplify $- 6 x - 16 \cdot 0$ .

Multiply $- 16$ by $0$ .

$- 6 x + 0 = - 30$

Add $- 6 x$ and $0$ .

$- 6 x = - 30$

Divide each term in $- 6 x = - 30$ by $- 6$ and simplify.

Divide each term in $- 6 x = - 30$ by $- 6$ .

$\frac{- 6 x}{- 6} = \frac{- 30}{- 6}$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 x}{- 6} = \frac{- 30}{- 6}$

Divide $x$ by $1$ .

$x = \frac{- 30}{- 6}$

Simplify the right side.

Divide $- 30$ by $- 6$ .

$x = 5$

Step 6

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

$(5, 0)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(5, 0)$

Equation Form:

$x = 5, y = 0$

Step 8

Solve by Addition/Elimination 3x+8y=15 , 2x-8y=10

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