Precalculus Examples

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

Step 1

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

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

$3 x + y - z = 4$

Step 2

Eliminate $x$ from the system.

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Step 2.1

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

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

$3 x + y - z = 4$

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 $(- 3) \cdot (x - 3 y + 4 z)$ .

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Step 2.2.1.1.1

Apply the distributive property.

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

$3 x + y - z = 4$

Step 2.2.1.1.2

Simplify.

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Step 2.2.1.1.2.1

Multiply $- 3$ by $- 3$ .

$- 3 x + 9 y - 3 (4 z) = (- 3) (- 3)$

$3 x + y - z = 4$

Step 2.2.1.1.2.2

Multiply $4$ by $- 3$ .

$- 3 x + 9 y - 12 z = (- 3) (- 3)$

$3 x + y - z = 4$

$- 3 x + 9 y - 12 z = (- 3) (- 3)$

$3 x + y - z = 4$

$- 3 x + 9 y - 12 z = (- 3) (- 3)$

$3 x + y - z = 4$

$- 3 x + 9 y - 12 z = (- 3) (- 3)$

$3 x + y - z = 4$

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

Multiply $- 3$ by $- 3$ .

$- 3 x + 9 y - 12 z = 9$

$3 x + y - z = 4$

$- 3 x + 9 y - 12 z = 9$

$3 x + y - z = 4$

$- 3 x + 9 y - 12 z = 9$

$3 x + y - z = 4$

Step 2.3

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

	$-$	$3$	$x$	$+$	$9$	$y$	$-$	$1$	$2$	$z$	$=$	$9$
$+$		$3$	$x$	$+$		$y$			$-$	$z$	$=$	$4$
				$1$	$0$	$y$	$-$	$1$	$3$	$z$	$=$	$1$	$3$

Step 2.4

The resultant equation has $x$ eliminated.

$10 y - 13 z = 13$

Step 3

Choose another two equations and eliminate $x$ .

$3 x + y - z = 4$

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

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.

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

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

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

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Step 4.2.1.1.1

Apply the distributive property.

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

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

Step 4.2.1.1.2

Simplify.

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Step 4.2.1.1.2.1

Multiply $3$ by $2$ .

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

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

Step 4.2.1.1.2.2

Multiply $- 1$ by $2$ .

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

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

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

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

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

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

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

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

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Multiply $2$ by $4$ .

$6 x + 2 y - 2 z = 8$

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

$6 x + 2 y - 2 z = 8$

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

Step 4.2.3

Simplify the left side.

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Step 4.2.3.1

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

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Step 4.2.3.1.1

Apply the distributive property.

$6 x + 2 y - 2 z = 8$

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

Step 4.2.3.1.2

Simplify.

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Step 4.2.3.1.2.1

Multiply $- 2$ by $3$ .

$6 x + 2 y - 2 z = 8$

$- 6 x + 3 (2 y) + 3 (3 z) = (3) (- 5)$

Step 4.2.3.1.2.2

Multiply $2$ by $3$ .

$6 x + 2 y - 2 z = 8$

$- 6 x + 6 y + 3 (3 z) = (3) (- 5)$

Step 4.2.3.1.2.3

Multiply $3$ by $3$ .

$6 x + 2 y - 2 z = 8$

$- 6 x + 6 y + 9 z = (3) (- 5)$

$6 x + 2 y - 2 z = 8$

$- 6 x + 6 y + 9 z = (3) (- 5)$

$6 x + 2 y - 2 z = 8$

$- 6 x + 6 y + 9 z = (3) (- 5)$

$6 x + 2 y - 2 z = 8$

$- 6 x + 6 y + 9 z = (3) (- 5)$

Step 4.2.4

Simplify the right side.

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Step 4.2.4.1

Multiply $3$ by $- 5$ .

$6 x + 2 y - 2 z = 8$

$- 6 x + 6 y + 9 z = - 15$

$6 x + 2 y - 2 z = 8$

$- 6 x + 6 y + 9 z = - 15$

$6 x + 2 y - 2 z = 8$

$- 6 x + 6 y + 9 z = - 15$

Step 4.3

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

		$6$	$x$	$+$	$2$	$y$	$-$	$2$	$z$	$=$			$8$
$+$	$-$	$6$	$x$	$+$	$6$	$y$	$+$	$9$	$z$	$=$	$-$	$1$	$5$
					$8$	$y$	$+$	$7$	$z$	$=$		$-$	$7$

Step 4.4

The resultant equation has $x$ eliminated.

$8 y + 7 z = - 7$

Step 5

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

$10 y - 13 z = 13$

$8 y + 7 z = - 7$

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.

$(- 4) \cdot (10 y - 13 z) = (- 4) (13)$

$(5) \cdot (8 y + 7 z) = (5) (- 7)$

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 $(- 4) \cdot (10 y - 13 z)$ .

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Step 6.2.1.1.1

Apply the distributive property.

$- 4 (10 y) - 4 (- 13 z) = (- 4) (13)$

$(5) \cdot (8 y + 7 z) = (5) (- 7)$

Step 6.2.1.1.2

Multiply.

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Step 6.2.1.1.2.1

Multiply $10$ by $- 4$ .

$- 40 y - 4 (- 13 z) = (- 4) (13)$

$(5) \cdot (8 y + 7 z) = (5) (- 7)$

Step 6.2.1.1.2.2

Multiply $- 13$ by $- 4$ .

$- 40 y + 52 z = (- 4) (13)$

$(5) \cdot (8 y + 7 z) = (5) (- 7)$

$- 40 y + 52 z = (- 4) (13)$

$(5) \cdot (8 y + 7 z) = (5) (- 7)$

$- 40 y + 52 z = (- 4) (13)$

$(5) \cdot (8 y + 7 z) = (5) (- 7)$

$- 40 y + 52 z = (- 4) (13)$

$(5) \cdot (8 y + 7 z) = (5) (- 7)$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

Multiply $- 4$ by $13$ .

$- 40 y + 52 z = - 52$

$(5) \cdot (8 y + 7 z) = (5) (- 7)$

$- 40 y + 52 z = - 52$

$(5) \cdot (8 y + 7 z) = (5) (- 7)$

Step 6.2.3

Simplify the left side.

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Step 6.2.3.1

Simplify $(5) \cdot (8 y + 7 z)$ .

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Step 6.2.3.1.1

Apply the distributive property.

$- 40 y + 52 z = - 52$

$5 (8 y) + 5 (7 z) = (5) (- 7)$

Step 6.2.3.1.2

Multiply.

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Step 6.2.3.1.2.1

Multiply $8$ by $5$ .

$- 40 y + 52 z = - 52$

$40 y + 5 (7 z) = (5) (- 7)$

Step 6.2.3.1.2.2

Multiply $7$ by $5$ .

$- 40 y + 52 z = - 52$

$40 y + 35 z = (5) (- 7)$

$- 40 y + 52 z = - 52$

$40 y + 35 z = (5) (- 7)$

$- 40 y + 52 z = - 52$

$40 y + 35 z = (5) (- 7)$

$- 40 y + 52 z = - 52$

$40 y + 35 z = (5) (- 7)$

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

Multiply $5$ by $- 7$ .

$- 40 y + 52 z = - 52$

$40 y + 35 z = - 35$

$- 40 y + 52 z = - 52$

$40 y + 35 z = - 35$

$- 40 y + 52 z = - 52$

$40 y + 35 z = - 35$

Step 6.3

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

	$-$	$4$	$0$	$y$	$+$	$5$	$2$	$z$	$=$	$-$	$5$	$2$
$+$		$4$	$0$	$y$	$+$	$3$	$5$	$z$	$=$	$-$	$3$	$5$
						$8$	$7$	$z$	$=$	$-$	$8$	$7$

Step 6.4

The resultant equation has $y$ eliminated.

$87 z = - 87$

Step 6.5

Divide each term in $87 z = - 87$ by $87$ and simplify.

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Step 6.5.1

Divide each term in $87 z = - 87$ by $87$ .

$\frac{87 z}{87} = \frac{- 87}{87}$

Step 6.5.2

Simplify the left side.

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Step 6.5.2.1

Cancel the common factor of $87$ .

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Step 6.5.2.1.1

Cancel the common factor.

$\frac{87 z}{87} = \frac{- 87}{87}$

Step 6.5.2.1.2

Divide $z$ by $1$ .

$z = \frac{- 87}{87}$

Step 6.5.3

Simplify the right side.

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Step 6.5.3.1

Divide $- 87$ by $87$ .

$z = - 1$

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.

$10 y - 13 \cdot - 1 = 13$

Step 7.2

Solve for $y$ .

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Step 7.2.1

Multiply $- 13$ by $- 1$ .

$10 y + 13 = 13$

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

Subtract $13$ from both sides of the equation.

$10 y = 13 - 13$

Step 7.2.2.2

Subtract $13$ from $13$ .

$10 y = 0$

Step 7.2.3

Divide each term in $10 y = 0$ by $10$ and simplify.

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Step 7.2.3.1

Divide each term in $10 y = 0$ by $10$ .

$\frac{10 y}{10} = \frac{0}{10}$

Step 7.2.3.2

Simplify the left side.

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Step 7.2.3.2.1

Cancel the common factor of $10$ .

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Step 7.2.3.2.1.1

Cancel the common factor.

$\frac{10 y}{10} = \frac{0}{10}$

Step 7.2.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{0}{10}$

Step 7.2.3.3

Simplify the right side.

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Step 7.2.3.3.1

Divide $0$ by $10$ .

$y = 0$

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.

$x - 3 \cdot 0 + 4 (- 1) = - 3$

Step 8.2

Solve for $x$ .

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Step 8.2.1

Simplify $x - 3 \cdot 0 + 4 (- 1)$ .

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Step 8.2.1.1

Simplify each term.

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Step 8.2.1.1.1

Multiply $- 3$ by $0$ .

$x + 0 + 4 (- 1) = - 3$

Step 8.2.1.1.2

Multiply $4$ by $- 1$ .

$x + 0 - 4 = - 3$

Step 8.2.1.2

Add $x$ and $0$ .

$x - 4 = - 3$

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 $4$ to both sides of the equation.

$x = - 3 + 4$

Step 8.2.2.2

Add $- 3$ and $4$ .

$x = 1$

Step 9

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

$(1, 0, - 1)$

Step 10

The result can be shown in multiple forms.

Point Form:

$(1, 0, - 1)$

Equation Form:

$x = 1, y = 0, z = - 1$

	$-$	$3$	$x$	$+$	$9$	$y$	$-$	$1$	$2$	$z$	$=$	$9$
$+$		$3$	$x$	$+$		$y$			$-$	$z$	$=$	$4$
				$1$	$0$	$y$	$-$	$1$	$3$	$z$	$=$	$1$	$3$

		$6$	$x$	$+$	$2$	$y$	$-$	$2$	$z$	$=$			$8$
$+$	$-$	$6$	$x$	$+$	$6$	$y$	$+$	$9$	$z$	$=$	$-$	$1$	$5$
					$8$	$y$	$+$	$7$	$z$	$=$		$-$	$7$

	$-$	$4$	$0$	$y$	$+$	$5$	$2$	$z$	$=$	$-$	$5$	$2$
$+$		$4$	$0$	$y$	$+$	$3$	$5$	$z$	$=$	$-$	$3$	$5$
						$8$	$7$	$z$	$=$	$-$	$8$	$7$

	$-$	$3$	$x$	$+$	$9$	$y$	$-$	$1$	$2$	$z$	$=$	$9$
$+$		$3$	$x$	$+$		$y$			$-$	$z$	$=$	$4$
				$1$	$0$	$y$	$-$	$1$	$3$	$z$	$=$	$1$	$3$

		$6$	$x$	$+$	$2$	$y$	$-$	$2$	$z$	$=$			$8$
$+$	$-$	$6$	$x$	$+$	$6$	$y$	$+$	$9$	$z$	$=$	$-$	$1$	$5$
					$8$	$y$	$+$	$7$	$z$	$=$		$-$	$7$

	$-$	$4$	$0$	$y$	$+$	$5$	$2$	$z$	$=$	$-$	$5$	$2$
$+$		$4$	$0$	$y$	$+$	$3$	$5$	$z$	$=$	$-$	$3$	$5$
						$8$	$7$	$z$	$=$	$-$	$8$	$7$

	$-$	$3$	$x$	$+$	$9$	$y$	$-$	$1$	$2$	$z$	$=$	$9$
$+$		$3$	$x$	$+$		$y$			$-$	$z$	$=$	$4$
				$1$	$0$	$y$	$-$	$1$	$3$	$z$	$=$	$1$	$3$

		$6$	$x$	$+$	$2$	$y$	$-$	$2$	$z$	$=$			$8$
$+$	$-$	$6$	$x$	$+$	$6$	$y$	$+$	$9$	$z$	$=$	$-$	$1$	$5$
					$8$	$y$	$+$	$7$	$z$	$=$		$-$	$7$

	$-$	$4$	$0$	$y$	$+$	$5$	$2$	$z$	$=$	$-$	$5$	$2$
$+$		$4$	$0$	$y$	$+$	$3$	$5$	$z$	$=$	$-$	$3$	$5$
						$8$	$7$	$z$	$=$	$-$	$8$	$7$