Solve by Addition/Elimination x+y=12 , x-y=4

$x + y = 12$ , $x - y = 4$

Step 1

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

$x + y = 12$

$(- 1) \cdot (x - y) = (- 1) (4)$

Step 2

Simplify.

Simplify the left side.

Simplify $(- 1) \cdot (x - y)$ .

Apply the distributive property.

$x + y = 12$

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

Rewrite $- 1 x$ as $- x$ .

$x + y = 12$

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

Multiply $- 1 (- y)$ .

Multiply $- 1$ by $- 1$ .

$x + y = 12$

$- x + 1 y = (- 1) (4)$

Multiply $y$ by $1$ .

$x + y = 12$

$- x + y = (- 1) (4)$

$x + y = 12$

$- x + y = (- 1) (4)$

$x + y = 12$

$- x + y = (- 1) (4)$

$x + y = 12$

$- x + y = (- 1) (4)$

Simplify the right side.

Multiply $- 1$ by $4$ .

$x + y = 12$

$- x + y = - 4$

$x + y = 12$

$- x + y = - 4$

$x + y = 12$

$- x + y = - 4$

Step 3

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

		$x$	$+$	$y$	$=$	$1$	$2$
$+$	$-$	$x$	$+$	$y$	$=$	$-$	$4$
			$2$	$y$	$=$		$8$

Step 4

Divide each term in $2 y = 8$ by $2$ and simplify.

Divide each term in $2 y = 8$ by $2$ .

$\frac{2 y}{2} = \frac{8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{8}{2}$

Divide $y$ by $1$ .

$y = \frac{8}{2}$

Simplify the right side.

Divide $8$ by $2$ .

$y = 4$

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 $x$ .

$x + 4 = 12$

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

Subtract $4$ from both sides of the equation.

$x = 12 - 4$

Subtract $4$ from $12$ .

$x = 8$

Step 6

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

$(8, 4)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(8, 4)$

Equation Form:

$x = 8, y = 4$

Step 8

Solve by Addition/Elimination x+y=12 , x-y=4

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