Solve by Addition/Elimination 2x+4y=15 2x+4y=13

$2 x + 4 y = 15$ $2 x + 4 y = 13$

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

$(- 1) \cdot (2 x + 4 y) = (- 1) (15)$

$2 x + 4 y = 13$

Simplify.

Simplify the left side.

Simplify $(- 1) \cdot (2 x + 4 y)$ .

Apply the distributive property.

$- 1 (2 x) - 1 (4 y) = (- 1) (15)$

$2 x + 4 y = 13$

Multiply.

Multiply $2$ by $- 1$ .

$- 2 x - 1 (4 y) = (- 1) (15)$

$2 x + 4 y = 13$

Multiply $4$ by $- 1$ .

$- 2 x - 4 y = (- 1) (15)$

$2 x + 4 y = 13$

$- 2 x - 4 y = (- 1) (15)$

$2 x + 4 y = 13$

$- 2 x - 4 y = (- 1) (15)$

$2 x + 4 y = 13$

$- 2 x - 4 y = (- 1) (15)$

$2 x + 4 y = 13$

Simplify the right side.

Multiply $- 1$ by $15$ .

$- 2 x - 4 y = - 15$

$2 x + 4 y = 13$

$- 2 x - 4 y = - 15$

$2 x + 4 y = 13$

$- 2 x - 4 y = - 15$

$2 x + 4 y = 13$

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

			$-$	$2$	$x$	$-$	$4$	$y$	$=$	$-$	$1$	$5$
$+$				$2$	$x$	$+$	$4$	$y$	$=$		$1$	$3$
								$0$	$=$		$-$	$2$

Since $0 \neq - 2$ , there are no solutions.

No solution

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