Precalculus Examples

$x + y - 3 z = - 8$ , $- x - 2 y + z = 4$ , $- x + y + 4 z = 7$

Step 1

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

$x + y - 3 z = - 8$

$- x - 2 y + z = 4$

Step 2

Eliminate $x$ from the system.

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Step 2.1

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

		$x$	$+$		$y$	$-$	$3$	$z$	$=$	$-$	$8$
$+$	$-$	$x$	$-$	$2$	$y$	$+$		$z$	$=$		$4$
				$-$	$y$	$-$	$2$	$z$	$=$	$-$	$4$

Step 2.2

The resultant equation has $x$ eliminated.

$- y - 2 z = - 4$

Step 3

Choose another two equations and eliminate $x$ .

$- x - 2 y + z = 4$

$- x + y + 4 z = 7$

Step 4

Eliminate $x$ from the system.

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Step 4.1

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

$- x - 2 y + z = 4$

$(- 1) \cdot (- x + y + 4 z) = (- 1) (7)$

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 $(- 1) \cdot (- x + y + 4 z)$ .

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Step 4.2.1.1.1

Apply the distributive property.

$- x - 2 y + z = 4$

$- 1 (- x) - 1 y - 1 (4 z) = (- 1) (7)$

Step 4.2.1.1.2

Simplify.

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Step 4.2.1.1.2.1

Multiply $- 1 (- x)$ .

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Step 4.2.1.1.2.1.1

Multiply $- 1$ by $- 1$ .

$- x - 2 y + z = 4$

$1 x - 1 y - 1 (4 z) = (- 1) (7)$

Step 4.2.1.1.2.1.2

Multiply $x$ by $1$ .

$- x - 2 y + z = 4$

$x - 1 y - 1 (4 z) = (- 1) (7)$

$- x - 2 y + z = 4$

$x - 1 y - 1 (4 z) = (- 1) (7)$

Step 4.2.1.1.2.2

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

$- x - 2 y + z = 4$

$x - y - 1 (4 z) = (- 1) (7)$

Step 4.2.1.1.2.3

Multiply $4$ by $- 1$ .

$- x - 2 y + z = 4$

$x - y - 4 z = (- 1) (7)$

$- x - 2 y + z = 4$

$x - y - 4 z = (- 1) (7)$

$- x - 2 y + z = 4$

$x - y - 4 z = (- 1) (7)$

$- x - 2 y + z = 4$

$x - y - 4 z = (- 1) (7)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Multiply $- 1$ by $7$ .

$- x - 2 y + z = 4$

$x - y - 4 z = - 7$

$- x - 2 y + z = 4$

$x - y - 4 z = - 7$

$- x - 2 y + z = 4$

$x - y - 4 z = - 7$

Step 4.3

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

	$-$	$x$	$-$	$2$	$y$	$+$		$z$	$=$		$4$
$+$		$x$		$-$	$y$	$-$	$4$	$z$	$=$	$-$	$7$
			$-$	$3$	$y$	$-$	$3$	$z$	$=$	$-$	$3$

Step 4.4

The resultant equation has $x$ eliminated.

$- 3 y - 3 z = - 3$

Step 5

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

$- y - 2 z = - 4$

$- 3 y - 3 z = - 3$

Step 6

Eliminate $y$ from the system.

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Step 6.1

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

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

$- 3 y - 3 z = - 3$

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 $(- 3) \cdot (- y - 2 z)$ .

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Step 6.2.1.1.1

Apply the distributive property.

$- 3 (- y) - 3 (- 2 z) = (- 3) (- 4)$

$- 3 y - 3 z = - 3$

Step 6.2.1.1.2

Multiply.

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Step 6.2.1.1.2.1

Multiply $- 1$ by $- 3$ .

$3 y - 3 (- 2 z) = (- 3) (- 4)$

$- 3 y - 3 z = - 3$

Step 6.2.1.1.2.2

Multiply $- 2$ by $- 3$ .

$3 y + 6 z = (- 3) (- 4)$

$- 3 y - 3 z = - 3$

$3 y + 6 z = (- 3) (- 4)$

$- 3 y - 3 z = - 3$

$3 y + 6 z = (- 3) (- 4)$

$- 3 y - 3 z = - 3$

$3 y + 6 z = (- 3) (- 4)$

$- 3 y - 3 z = - 3$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

Multiply $- 3$ by $- 4$ .

$3 y + 6 z = 12$

$- 3 y - 3 z = - 3$

$3 y + 6 z = 12$

$- 3 y - 3 z = - 3$

$3 y + 6 z = 12$

$- 3 y - 3 z = - 3$

Step 6.3

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

		$3$	$y$	$+$	$6$	$z$	$=$	$1$	$2$
$+$	$-$	$3$	$y$	$-$	$3$	$z$	$=$	$-$	$3$
					$3$	$z$	$=$		$9$

Step 6.4

The resultant equation has $y$ eliminated.

$3 z = 9$

Step 6.5

Divide each term in $3 z = 9$ by $3$ and simplify.

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Step 6.5.1

Divide each term in $3 z = 9$ by $3$ .

$\frac{3 z}{3} = \frac{9}{3}$

Step 6.5.2

Simplify the left side.

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Step 6.5.2.1

Cancel the common factor of $3$ .

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Step 6.5.2.1.1

Cancel the common factor.

$\frac{3 z}{3} = \frac{9}{3}$

Step 6.5.2.1.2

Divide $z$ by $1$ .

$z = \frac{9}{3}$

Step 6.5.3

Simplify the right side.

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Step 6.5.3.1

Divide $9$ by $3$ .

$z = 3$

Step 7

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

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Step 7.1

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

$- y - 2 \cdot 3 = - 4$

Step 7.2

Solve for $y$ .

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Step 7.2.1

Multiply $- 2$ by $3$ .

$- y - 6 = - 4$

Step 7.2.2

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

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Step 7.2.2.1

Add $6$ to both sides of the equation.

$- y = - 4 + 6$

Step 7.2.2.2

Add $- 4$ and $6$ .

$- y = 2$

Step 7.2.3

Divide each term in $- y = 2$ by $- 1$ and simplify.

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Step 7.2.3.1

Divide each term in $- y = 2$ by $- 1$ .

$\frac{- y}{- 1} = \frac{2}{- 1}$

Step 7.2.3.2

Simplify the left side.

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Step 7.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{2}{- 1}$

Step 7.2.3.2.2

Divide $y$ by $1$ .

$y = \frac{2}{- 1}$

Step 7.2.3.3

Simplify the right side.

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Step 7.2.3.3.1

Divide $2$ by $- 1$ .

$y = - 2$

Step 8

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

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Step 8.1

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

$x - 2 - 3 \cdot 3 = - 8$

Step 8.2

Solve for $x$ .

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Step 8.2.1

Simplify $x - 2 - 3 \cdot 3$ .

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Step 8.2.1.1

Multiply $- 3$ by $3$ .

$x - 2 - 9 = - 8$

Step 8.2.1.2

Subtract $9$ from $- 2$ .

$x - 11 = - 8$

Step 8.2.2

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

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Step 8.2.2.1

Add $11$ to both sides of the equation.

$x = - 8 + 11$

Step 8.2.2.2

Add $- 8$ and $11$ .

$x = 3$

Step 9

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

$(3, - 2, 3)$

Step 10

The result can be shown in multiple forms.

Point Form:

$(3, - 2, 3)$

Equation Form:

$x = 3, y = - 2, z = 3$

		$x$	$+$		$y$	$-$	$3$	$z$	$=$	$-$	$8$
$+$	$-$	$x$	$-$	$2$	$y$	$+$		$z$	$=$		$4$
				$-$	$y$	$-$	$2$	$z$	$=$	$-$	$4$

	$-$	$x$	$-$	$2$	$y$	$+$		$z$	$=$		$4$
$+$		$x$		$-$	$y$	$-$	$4$	$z$	$=$	$-$	$7$
			$-$	$3$	$y$	$-$	$3$	$z$	$=$	$-$	$3$

		$3$	$y$	$+$	$6$	$z$	$=$	$1$	$2$
$+$	$-$	$3$	$y$	$-$	$3$	$z$	$=$	$-$	$3$
					$3$	$z$	$=$		$9$

		$x$	$+$		$y$	$-$	$3$	$z$	$=$	$-$	$8$
$+$	$-$	$x$	$-$	$2$	$y$	$+$		$z$	$=$		$4$
				$-$	$y$	$-$	$2$	$z$	$=$	$-$	$4$

	$-$	$x$	$-$	$2$	$y$	$+$		$z$	$=$		$4$
$+$		$x$		$-$	$y$	$-$	$4$	$z$	$=$	$-$	$7$
			$-$	$3$	$y$	$-$	$3$	$z$	$=$	$-$	$3$

		$3$	$y$	$+$	$6$	$z$	$=$	$1$	$2$
$+$	$-$	$3$	$y$	$-$	$3$	$z$	$=$	$-$	$3$
					$3$	$z$	$=$		$9$

		$x$	$+$		$y$	$-$	$3$	$z$	$=$	$-$	$8$
$+$	$-$	$x$	$-$	$2$	$y$	$+$		$z$	$=$		$4$
				$-$	$y$	$-$	$2$	$z$	$=$	$-$	$4$

	$-$	$x$	$-$	$2$	$y$	$+$		$z$	$=$		$4$
$+$		$x$		$-$	$y$	$-$	$4$	$z$	$=$	$-$	$7$
			$-$	$3$	$y$	$-$	$3$	$z$	$=$	$-$	$3$

		$3$	$y$	$+$	$6$	$z$	$=$	$1$	$2$
$+$	$-$	$3$	$y$	$-$	$3$	$z$	$=$	$-$	$3$
					$3$	$z$	$=$		$9$