Solve by Addition/Elimination x+3y-z=2 x+y-z=0 3x+2y-3z=-1

$x + 3 y - z = 2$ $x + y - z = 0$ $3 x + 2 y - 3 z = - 1$

Choose two equations and eliminate one variable. In this case, eliminate $x$ .

$x + 3 y - z = 2$

$x + y - z = 0$

Eliminate $x$ from the system.

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

$x + 3 y - z = 2$

$(- 1) \cdot (x + y - z) = (- 1) (0)$

Simplify.

Simplify the left side.

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

Apply the distributive property.

$x + 3 y - z = 2$

$- 1 x - 1 y - 1 (- z) = (- 1) (0)$

Simplify.

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

$x + 3 y - z = 2$

$- x - 1 y - 1 (- z) = (- 1) (0)$

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

$x + 3 y - z = 2$

$- x - y - 1 (- z) = (- 1) (0)$

Multiply $- 1 (- z)$ .

Multiply $- 1$ by $- 1$ .

$x + 3 y - z = 2$

$- x - y + 1 z = (- 1) (0)$

Multiply $z$ by $1$ .

$x + 3 y - z = 2$

$- x - y + z = (- 1) (0)$

$x + 3 y - z = 2$

$- x - y + z = (- 1) (0)$

$x + 3 y - z = 2$

$- x - y + z = (- 1) (0)$

$x + 3 y - z = 2$

$- x - y + z = (- 1) (0)$

$x + 3 y - z = 2$

$- x - y + z = (- 1) (0)$

Simplify the right side.

Multiply $- 1$ by $0$ .

$x + 3 y - z = 2$

$- x - y + z = 0$

$x + 3 y - z = 2$

$- x - y + z = 0$

$x + 3 y - z = 2$

$- x - y + z = 0$

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

		$x$	$+$	$3$	$y$	$-$	$z$	$=$	$2$
$+$	$-$	$x$		$-$	$y$	$+$	$z$	$=$	$0$
				$2$	$y$			$=$	$2$

The resultant equation has $x$ eliminated.

$2 y = 2$

Choose another two equations and eliminate $x$ .

$x + y - z = 0$

$3 x + 2 y - 3 z = - 1$

Eliminate $x$ from the system.

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

$(- 3) \cdot (x + y - z) = (- 3) (0)$

$3 x + 2 y - 3 z = - 1$

Simplify.

Simplify the left side.

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

Apply the distributive property.

$- 3 x - 3 y - 3 (- z) = (- 3) (0)$

$3 x + 2 y - 3 z = - 1$

Multiply $- 1$ by $- 3$ .

$- 3 x - 3 y + 3 z = (- 3) (0)$

$3 x + 2 y - 3 z = - 1$

$- 3 x - 3 y + 3 z = (- 3) (0)$

$3 x + 2 y - 3 z = - 1$

$- 3 x - 3 y + 3 z = (- 3) (0)$

$3 x + 2 y - 3 z = - 1$

Simplify the right side.

Multiply $- 3$ by $0$ .

$- 3 x - 3 y + 3 z = 0$

$3 x + 2 y - 3 z = - 1$

$- 3 x - 3 y + 3 z = 0$

$3 x + 2 y - 3 z = - 1$

$- 3 x - 3 y + 3 z = 0$

$3 x + 2 y - 3 z = - 1$

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

	$-$	$3$	$x$	$-$	$3$	$y$	$+$	$3$	$z$	$=$		$0$
$+$		$3$	$x$	$+$	$2$	$y$	$-$	$3$	$z$	$=$	$-$	$1$
					$-$	$y$				$=$	$-$	$1$

The resultant equation has $x$ eliminated.

$- y = - 1$

Take the resultant equations and eliminate another variable. In this case, eliminate $y$ .

$2 y = 2$

$- y = - 1$

Eliminate $y$ from the system.

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

$2 y = 2$

$(2) \cdot (- y) = (2) (- 1)$

Simplify.

Simplify the left side.

Multiply $- 1$ by $2$ .

$2 y = 2$

$- 2 y = (2) (- 1)$

$2 y = 2$

$- 2 y = (2) (- 1)$

Simplify the right side.

Multiply $2$ by $- 1$ .

$2 y = 2$

$- 2 y = - 2$

$2 y = 2$

$- 2 y = - 2$

$2 y = 2$

$- 2 y = - 2$

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

		$2$	$y$	$=$		$2$
$+$	$-$	$2$	$y$	$=$	$-$	$2$
			$0$	$=$		$0$

The resultant equation has $y$ eliminated.

$0 = 0$

Because the resultant equation includes no variables and is true, the system of equations has an infinite number of solutions.

Infinite number of solutions

Solve by Addition/Elimination x+3y-z=2 x+y-z=0 3x+2y-3z=-1

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