Precalculus Examples

$5 x - 3 y + 2 z = 4$ , $- 9 x + 5 y - 4 z = - 12$ , $- 3 x + y - 2 z = - 12$

Step 1

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

$5 x - 3 y + 2 z = 4$

$- 9 x + 5 y - 4 z = - 12$

Step 2

Eliminate $z$ from the system.

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Step 2.1

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

$(2) \cdot (5 x - 3 y + 2 z) = (2) (4)$

$- 9 x + 5 y - 4 z = - 12$

Step 2.2

Simplify.

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Step 2.2.1

Simplify the left side.

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Step 2.2.1.1

Simplify $(2) \cdot (5 x - 3 y + 2 z)$ .

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Step 2.2.1.1.1

Apply the distributive property.

$2 (5 x) + 2 (- 3 y) + 2 (2 z) = (2) (4)$

$- 9 x + 5 y - 4 z = - 12$

Step 2.2.1.1.2

Simplify.

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Step 2.2.1.1.2.1

Multiply $5$ by $2$ .

$10 x + 2 (- 3 y) + 2 (2 z) = (2) (4)$

$- 9 x + 5 y - 4 z = - 12$

Step 2.2.1.1.2.2

Multiply $- 3$ by $2$ .

$10 x - 6 y + 2 (2 z) = (2) (4)$

$- 9 x + 5 y - 4 z = - 12$

Step 2.2.1.1.2.3

Multiply $2$ by $2$ .

$10 x - 6 y + 4 z = (2) (4)$

$- 9 x + 5 y - 4 z = - 12$

$10 x - 6 y + 4 z = (2) (4)$

$- 9 x + 5 y - 4 z = - 12$

$10 x - 6 y + 4 z = (2) (4)$

$- 9 x + 5 y - 4 z = - 12$

$10 x - 6 y + 4 z = (2) (4)$

$- 9 x + 5 y - 4 z = - 12$

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

Multiply $2$ by $4$ .

$10 x - 6 y + 4 z = 8$

$- 9 x + 5 y - 4 z = - 12$

$10 x - 6 y + 4 z = 8$

$- 9 x + 5 y - 4 z = - 12$

$10 x - 6 y + 4 z = 8$

$- 9 x + 5 y - 4 z = - 12$

Step 2.3

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

	$1$	$0$	$x$	$-$	$6$	$y$	$+$	$4$	$z$	$=$			$8$
$+$	$-$	$9$	$x$	$+$	$5$	$y$	$-$	$4$	$z$	$=$	$-$	$1$	$2$
			$x$		$-$	$y$				$=$		$-$	$4$

Step 2.4

The resultant equation has $z$ eliminated.

$x - y = - 4$

Step 3

Choose another two equations and eliminate $z$ .

$- 9 x + 5 y - 4 z = - 12$

$- 3 x + y - 2 z = - 12$

Step 4

Eliminate $z$ from the system.

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Step 4.1

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

$- 9 x + 5 y - 4 z = - 12$

$(- 2) \cdot (- 3 x + y - 2 z) = (- 2) (- 12)$

Step 4.2

Simplify.

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Step 4.2.1

Simplify the left side.

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Step 4.2.1.1

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

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Step 4.2.1.1.1

Apply the distributive property.

$- 9 x + 5 y - 4 z = - 12$

$- 2 (- 3 x) - 2 y - 2 (- 2 z) = (- 2) (- 12)$

Step 4.2.1.1.2

Simplify.

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Step 4.2.1.1.2.1

Multiply $- 3$ by $- 2$ .

$- 9 x + 5 y - 4 z = - 12$

$6 x - 2 y - 2 (- 2 z) = (- 2) (- 12)$

Step 4.2.1.1.2.2

Multiply $- 2$ by $- 2$ .

$- 9 x + 5 y - 4 z = - 12$

$6 x - 2 y + 4 z = (- 2) (- 12)$

$- 9 x + 5 y - 4 z = - 12$

$6 x - 2 y + 4 z = (- 2) (- 12)$

$- 9 x + 5 y - 4 z = - 12$

$6 x - 2 y + 4 z = (- 2) (- 12)$

$- 9 x + 5 y - 4 z = - 12$

$6 x - 2 y + 4 z = (- 2) (- 12)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Multiply $- 2$ by $- 12$ .

$- 9 x + 5 y - 4 z = - 12$

$6 x - 2 y + 4 z = 24$

$- 9 x + 5 y - 4 z = - 12$

$6 x - 2 y + 4 z = 24$

$- 9 x + 5 y - 4 z = - 12$

$6 x - 2 y + 4 z = 24$

Step 4.3

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

	$-$	$9$	$x$	$+$	$5$	$y$	$-$	$4$	$z$	$=$	$-$	$1$	$2$
$+$		$6$	$x$	$-$	$2$	$y$	$+$	$4$	$z$	$=$		$2$	$4$
	$-$	$3$	$x$	$+$	$3$	$y$				$=$		$1$	$2$

Step 4.4

The resultant equation has $z$ eliminated.

$- 3 x + 3 y = 12$

Step 5

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

$x - y = - 4$

$- 3 x + 3 y = 12$

Step 6

Eliminate $x$ from the system.

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Step 6.1

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

$(3) \cdot (x - y) = (3) (- 4)$

$- 3 x + 3 y = 12$

Step 6.2

Simplify.

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Step 6.2.1

Simplify the left side.

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Step 6.2.1.1

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

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Step 6.2.1.1.1

Apply the distributive property.

$3 x + 3 (- y) = (3) (- 4)$

$- 3 x + 3 y = 12$

Step 6.2.1.1.2

Multiply $- 1$ by $3$ .

$3 x - 3 y = (3) (- 4)$

$- 3 x + 3 y = 12$

$3 x - 3 y = (3) (- 4)$

$- 3 x + 3 y = 12$

$3 x - 3 y = (3) (- 4)$

$- 3 x + 3 y = 12$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

Multiply $3$ by $- 4$ .

$3 x - 3 y = - 12$

$- 3 x + 3 y = 12$

$3 x - 3 y = - 12$

$- 3 x + 3 y = 12$

$3 x - 3 y = - 12$

$- 3 x + 3 y = 12$

Step 6.3

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

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

Step 6.4

The resultant equation has $x$ eliminated.

$0 = 0$

Step 7

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

Infinite number of solutions

	$1$	$0$	$x$	$-$	$6$	$y$	$+$	$4$	$z$	$=$			$8$
$+$	$-$	$9$	$x$	$+$	$5$	$y$	$-$	$4$	$z$	$=$	$-$	$1$	$2$
			$x$		$-$	$y$				$=$		$-$	$4$

	$-$	$9$	$x$	$+$	$5$	$y$	$-$	$4$	$z$	$=$	$-$	$1$	$2$
$+$		$6$	$x$	$-$	$2$	$y$	$+$	$4$	$z$	$=$		$2$	$4$
	$-$	$3$	$x$	$+$	$3$	$y$				$=$		$1$	$2$

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

	$1$	$0$	$x$	$-$	$6$	$y$	$+$	$4$	$z$	$=$			$8$
$+$	$-$	$9$	$x$	$+$	$5$	$y$	$-$	$4$	$z$	$=$	$-$	$1$	$2$
			$x$		$-$	$y$				$=$		$-$	$4$

	$-$	$9$	$x$	$+$	$5$	$y$	$-$	$4$	$z$	$=$	$-$	$1$	$2$
$+$		$6$	$x$	$-$	$2$	$y$	$+$	$4$	$z$	$=$		$2$	$4$
	$-$	$3$	$x$	$+$	$3$	$y$				$=$		$1$	$2$

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

	$1$	$0$	$x$	$-$	$6$	$y$	$+$	$4$	$z$	$=$			$8$
$+$	$-$	$9$	$x$	$+$	$5$	$y$	$-$	$4$	$z$	$=$	$-$	$1$	$2$
			$x$		$-$	$y$				$=$		$-$	$4$

	$-$	$9$	$x$	$+$	$5$	$y$	$-$	$4$	$z$	$=$	$-$	$1$	$2$
$+$		$6$	$x$	$-$	$2$	$y$	$+$	$4$	$z$	$=$		$2$	$4$
	$-$	$3$	$x$	$+$	$3$	$y$				$=$		$1$	$2$

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