Solve by Addition/Elimination y+z=4 y-z=8

$y + z = 4$ $y - z = 8$

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

$y + z = 4$

$(- 1) \cdot (y - z) = (- 1) (8)$

Simplify.

Simplify the left side.

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

Apply the distributive property.

$y + z = 4$

$- 1 y - 1 (- z) = (- 1) (8)$

Rewrite $- 1 y$ as $- y$ .

$y + z = 4$

$- y - 1 (- z) = (- 1) (8)$

Multiply $- 1 (- z)$ .

Multiply $- 1$ by $- 1$ .

$y + z = 4$

$- y + 1 z = (- 1) (8)$

Multiply $z$ by $1$ .

$y + z = 4$

$- y + z = (- 1) (8)$

$y + z = 4$

$- y + z = (- 1) (8)$

$y + z = 4$

$- y + z = (- 1) (8)$

$y + z = 4$

$- y + z = (- 1) (8)$

Simplify the right side.

Multiply $- 1$ by $8$ .

$y + z = 4$

$- y + z = - 8$

$y + z = 4$

$- y + z = - 8$

$y + z = 4$

$- y + z = - 8$

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

		$y$	$+$	$z$	$=$		$4$
$+$	$-$	$y$	$+$	$z$	$=$	$-$	$8$
			$2$	$z$	$=$	$-$	$4$

Divide each term in $2 z = - 4$ by $2$ and simplify.

Divide each term in $2 z = - 4$ by $2$ .

$\frac{2 z}{2} = \frac{- 4}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 z}{2} = \frac{- 4}{2}$

Divide $z$ by $1$ .

$z = \frac{- 4}{2}$

Simplify the right side.

Divide $- 4$ by $2$ .

$z = - 2$

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

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

$y - 2 = 4$

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

Add $2$ to both sides of the equation.

$y = 4 + 2$

Add $4$ and $2$ .

$y = 6$

This is the final solution to the independent system of equations.

$z = - 2$

$y = 6$

Solve by Addition/Elimination y+z=4 y-z=8

Related Questions

Support LCMGCF.com by disabling your adblocker.