Precalculus Examples

$- 2 x + y - 2 z = - 15$ , $- x + y - z = - 11$ , $z = - 5$

Step 1

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

$- 2 x + y - 2 z = - 15$

$- x + y - z = - 11$

Step 2

Eliminate $y$ from the system.

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Step 2.1

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

$- 2 x + y - 2 z = - 15$

$(- 1) \cdot (- x + y - z) = (- 1) (- 11)$

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

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Step 2.2.1.1.1

Apply the distributive property.

$- 2 x + y - 2 z = - 15$

$- 1 (- x) - 1 y - 1 (- z) = (- 1) (- 11)$

Step 2.2.1.1.2

Simplify.

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Step 2.2.1.1.2.1

Multiply $- 1 (- x)$ .

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Step 2.2.1.1.2.1.1

Multiply $- 1$ by $- 1$ .

$- 2 x + y - 2 z = - 15$

$1 x - 1 y - 1 (- z) = (- 1) (- 11)$

Step 2.2.1.1.2.1.2

Multiply $x$ by $1$ .

$- 2 x + y - 2 z = - 15$

$x - 1 y - 1 (- z) = (- 1) (- 11)$

$- 2 x + y - 2 z = - 15$

$x - 1 y - 1 (- z) = (- 1) (- 11)$

Step 2.2.1.1.2.2

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

$- 2 x + y - 2 z = - 15$

$x - y - 1 (- z) = (- 1) (- 11)$

Step 2.2.1.1.2.3

Multiply $- 1 (- z)$ .

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Step 2.2.1.1.2.3.1

Multiply $- 1$ by $- 1$ .

$- 2 x + y - 2 z = - 15$

$x - y + 1 z = (- 1) (- 11)$

Step 2.2.1.1.2.3.2

Multiply $z$ by $1$ .

$- 2 x + y - 2 z = - 15$

$x - y + z = (- 1) (- 11)$

$- 2 x + y - 2 z = - 15$

$x - y + z = (- 1) (- 11)$

$- 2 x + y - 2 z = - 15$

$x - y + z = (- 1) (- 11)$

$- 2 x + y - 2 z = - 15$

$x - y + z = (- 1) (- 11)$

$- 2 x + y - 2 z = - 15$

$x - y + z = (- 1) (- 11)$

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

Multiply $- 1$ by $- 11$ .

$- 2 x + y - 2 z = - 15$

$x - y + z = 11$

$- 2 x + y - 2 z = - 15$

$x - y + z = 11$

$- 2 x + y - 2 z = - 15$

$x - y + z = 11$

Step 2.3

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

	$-$	$2$	$x$	$+$	$y$	$-$	$2$	$z$	$=$	$-$	$1$	$5$
$+$			$x$	$-$	$y$	$+$		$z$	$=$		$1$	$1$
		$-$	$x$				$-$	$z$	$=$		$-$	$4$

Step 2.4

The resultant equation has $y$ eliminated.

$- x - z = - 4$

Step 3

Take the resultant equation and the third original equation and eliminate another variable. In this case, eliminate $z$ .

$- x - z = - 4$

$z = - 5$

Step 4

Eliminate $z$ from the system.

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Step 4.1

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

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

Step 4.2

The resultant equation has $z$ eliminated.

$- x = - 9$

Step 4.3

Divide each term in $- x = - 9$ by $- 1$ and simplify.

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Step 4.3.1

Divide each term in $- x = - 9$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 9}{- 1}$

Step 4.3.2

Simplify the left side.

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Step 4.3.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 9}{- 1}$

Step 4.3.2.2

Divide $x$ by $1$ .

$x = \frac{- 9}{- 1}$

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

Divide $- 9$ by $- 1$ .

$x = 9$

Step 5

Substitute the value of $x$ into an equation with $y$ eliminated already and solve for the remaining variable.

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Step 5.1

Substitute the value of $x$ into an equation with $y$ eliminated already.

$- (9) - z = - 4$

Step 5.2

Solve for $z$ .

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Step 5.2.1

Multiply $- 1$ by $9$ .

$- 9 - z = - 4$

Step 5.2.2

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

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Step 5.2.2.1

Add $9$ to both sides of the equation.

$- z = - 4 + 9$

Step 5.2.2.2

Add $- 4$ and $9$ .

$- z = 5$

Step 5.2.3

Divide each term in $- z = 5$ by $- 1$ and simplify.

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Step 5.2.3.1

Divide each term in $- z = 5$ by $- 1$ .

$\frac{- z}{- 1} = \frac{5}{- 1}$

Step 5.2.3.2

Simplify the left side.

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Step 5.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{z}{1} = \frac{5}{- 1}$

Step 5.2.3.2.2

Divide $z$ by $1$ .

$z = \frac{5}{- 1}$

Step 5.2.3.3

Simplify the right side.

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Step 5.2.3.3.1

Divide $5$ by $- 1$ .

$z = - 5$

Step 6

Substitute the value of each known variable into one of the initial equations and solve for the last variable.

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Step 6.1

Substitute the value of each known variable into one of the initial equations.

$- 2 \cdot 9 + y - 2 \cdot - 5 = - 15$

Step 6.2

Solve for $y$ .

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Step 6.2.1

Simplify $- 2 \cdot 9 + y - 2 \cdot - 5$ .

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Step 6.2.1.1

Simplify each term.

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Step 6.2.1.1.1

Multiply $- 2$ by $9$ .

$- 18 + y - 2 \cdot - 5 = - 15$

Step 6.2.1.1.2

Multiply $- 2$ by $- 5$ .

$- 18 + y + 10 = - 15$

Step 6.2.1.2

Add $- 18$ and $10$ .

$y - 8 = - 15$

Step 6.2.2

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

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Step 6.2.2.1

Add $8$ to both sides of the equation.

$y = - 15 + 8$

Step 6.2.2.2

Add $- 15$ and $8$ .

$y = - 7$

Step 7

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

$(9, - 7, - 5)$

Step 8

The result can be shown in multiple forms.

Point Form:

$(9, - 7, - 5)$

Equation Form:

$x = 9, y = - 7, z = - 5$

	$-$	$2$	$x$	$+$	$y$	$-$	$2$	$z$	$=$	$-$	$1$	$5$
$+$			$x$	$-$	$y$	$+$		$z$	$=$		$1$	$1$
		$-$	$x$				$-$	$z$	$=$		$-$	$4$

	$-$	$2$	$x$	$+$	$y$	$-$	$2$	$z$	$=$	$-$	$1$	$5$
$+$			$x$	$-$	$y$	$+$		$z$	$=$		$1$	$1$
		$-$	$x$				$-$	$z$	$=$		$-$	$4$

	$-$	$2$	$x$	$+$	$y$	$-$	$2$	$z$	$=$	$-$	$1$	$5$
$+$			$x$	$-$	$y$	$+$		$z$	$=$		$1$	$1$
		$-$	$x$				$-$	$z$	$=$		$-$	$4$