Precalculus Examples

$4 x + y - 2 z = 0$ , $2 x - 3 y + 3 z = 9$ , $- 6 x - 2 y + z = 0$

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

$4 x + y - 2 z = 0$

$2 x - 3 y + 3 z = 9$

Eliminate $y$ from the system.

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

$(3) \cdot (4 x + y - 2 z) = (3) (0)$

$2 x - 3 y + 3 z = 9$

Simplify.

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

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Apply the distributive property.

$3 (4 x) + 3 y + 3 (- 2 z) = (3) (0)$

$2 x - 3 y + 3 z = 9$

Simplify.

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Multiply $4$ by $3$ .

$12 x + 3 y + 3 (- 2 z) = (3) (0)$

$2 x - 3 y + 3 z = 9$

Multiply $- 2$ by $3$ .

$12 x + 3 y - 6 z = (3) (0)$

$2 x - 3 y + 3 z = 9$

$12 x + 3 y - 6 z = (3) (0)$

$2 x - 3 y + 3 z = 9$

$12 x + 3 y - 6 z = (3) (0)$

$2 x - 3 y + 3 z = 9$

Multiply $3$ by $0$ .

$12 x + 3 y - 6 z = 0$

$2 x - 3 y + 3 z = 9$

$12 x + 3 y - 6 z = 0$

$2 x - 3 y + 3 z = 9$

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

	$1$	$2$	$x$	$+$	$3$	$y$	$-$	$6$	$z$	$=$	$0$
$+$		$2$	$x$	$-$	$3$	$y$	$+$	$3$	$z$	$=$	$9$
	$1$	$4$	$x$				$-$	$3$	$z$	$=$	$9$

The resultant equation has $y$ eliminated.

$14 x - 3 z = 9$

Choose another two equations and eliminate $y$ .

$2 x - 3 y + 3 z = 9$

$- 6 x - 2 y + z = 0$

Eliminate $y$ from the system.

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

$(- 2) \cdot (2 x - 3 y + 3 z) = (- 2) (9)$

$(3) \cdot (- 6 x - 2 y + z) = (3) (0)$

Simplify.

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

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Apply the distributive property.

$- 2 (2 x) - 2 (- 3 y) - 2 (3 z) = (- 2) (9)$

$(3) \cdot (- 6 x - 2 y + z) = (3) (0)$

Simplify.

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Multiply $2$ by $- 2$ .

$- 4 x - 2 (- 3 y) - 2 (3 z) = (- 2) (9)$

$(3) \cdot (- 6 x - 2 y + z) = (3) (0)$

Multiply $- 3$ by $- 2$ .

$- 4 x + 6 y - 2 (3 z) = (- 2) (9)$

$(3) \cdot (- 6 x - 2 y + z) = (3) (0)$

Multiply $3$ by $- 2$ .

$- 4 x + 6 y - 6 z = (- 2) (9)$

$(3) \cdot (- 6 x - 2 y + z) = (3) (0)$

$- 4 x + 6 y - 6 z = (- 2) (9)$

$(3) \cdot (- 6 x - 2 y + z) = (3) (0)$

$- 4 x + 6 y - 6 z = (- 2) (9)$

$(3) \cdot (- 6 x - 2 y + z) = (3) (0)$

Multiply $- 2$ by $9$ .

$- 4 x + 6 y - 6 z = - 18$

$(3) \cdot (- 6 x - 2 y + z) = (3) (0)$

Simplify $(3) \cdot (- 6 x - 2 y + z)$ .

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Apply the distributive property.

$- 4 x + 6 y - 6 z = - 18$

$3 (- 6 x) + 3 (- 2 y) + 3 z = (3) (0)$

Simplify.

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Multiply $- 6$ by $3$ .

$- 4 x + 6 y - 6 z = - 18$

$- 18 x + 3 (- 2 y) + 3 z = (3) (0)$

Multiply $- 2$ by $3$ .

$- 4 x + 6 y - 6 z = - 18$

$- 18 x - 6 y + 3 z = (3) (0)$

$- 4 x + 6 y - 6 z = - 18$

$- 18 x - 6 y + 3 z = (3) (0)$

$- 4 x + 6 y - 6 z = - 18$

$- 18 x - 6 y + 3 z = (3) (0)$

Multiply $3$ by $0$ .

$- 4 x + 6 y - 6 z = - 18$

$- 18 x - 6 y + 3 z = 0$

$- 4 x + 6 y - 6 z = - 18$

$- 18 x - 6 y + 3 z = 0$

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

		$-$	$4$	$x$	$+$	$6$	$y$	$-$	$6$	$z$	$=$	$-$	$1$	$8$
$+$	$-$	$1$	$8$	$x$	$-$	$6$	$y$	$+$	$3$	$z$	$=$			$0$
	$-$	$2$	$2$	$x$				$-$	$3$	$z$	$=$	$-$	$1$	$8$

The resultant equation has $y$ eliminated.

$- 22 x - 3 z = - 18$

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

$14 x - 3 z = 9$

$- 22 x - 3 z = - 18$

Eliminate $z$ from the system.

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

$(- 1) \cdot (14 x - 3 z) = (- 1) (9)$

$- 22 x - 3 z = - 18$

Simplify.

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Simplify $(- 1) \cdot (14 x - 3 z)$ .

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Apply the distributive property.

$- 1 (14 x) - 1 (- 3 z) = (- 1) (9)$

$- 22 x - 3 z = - 18$

Multiply.

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Multiply $14$ by $- 1$ .

$- 14 x - 1 (- 3 z) = (- 1) (9)$

$- 22 x - 3 z = - 18$

Multiply $- 3$ by $- 1$ .

$- 14 x + 3 z = (- 1) (9)$

$- 22 x - 3 z = - 18$

$- 14 x + 3 z = (- 1) (9)$

$- 22 x - 3 z = - 18$

$- 14 x + 3 z = (- 1) (9)$

$- 22 x - 3 z = - 18$

Multiply $- 1$ by $9$ .

$- 14 x + 3 z = - 9$

$- 22 x - 3 z = - 18$

$- 14 x + 3 z = - 9$

$- 22 x - 3 z = - 18$

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

	$-$	$1$	$4$	$x$	$+$	$3$	$z$	$=$		$-$	$9$
$+$	$-$	$2$	$2$	$x$	$-$	$3$	$z$	$=$	$-$	$1$	$8$
	$-$	$3$	$6$	$x$				$=$	$-$	$2$	$7$

The resultant equation has $z$ eliminated.

$- 36 x = - 27$

Divide each term by $- 36$ and simplify.

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Divide each term in $- 36 x = - 27$ by $- 36$ .

$\frac{- 36 x}{- 36} = \frac{- 27}{- 36}$

Cancel the common factor of $- 36$ .

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Cancel the common factor.

$\frac{- 36 x}{- 36} = \frac{- 27}{- 36}$

Divide $x$ by $1$ .

$x = \frac{- 27}{- 36}$

Cancel the common factor of $- 27$ and $- 36$ .

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Factor $- 9$ out of $- 27$ .

$x = \frac{- 9 (3)}{- 36}$

Cancel the common factors.

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Factor $- 9$ out of $- 36$ .

$x = \frac{- 9 \cdot 3}{- 9 \cdot 4}$

Cancel the common factor.

$x = \frac{- 9 \cdot 3}{- 9 \cdot 4}$

Rewrite the expression.

$x = \frac{3}{4}$

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

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

$14 (\frac{3}{4}) - 3 z = 9$

Solve for $z$ .

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Simplify each term.

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Cancel the common factor of $2$ .

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Factor $2$ out of $14$ .

$2 (7) \frac{3}{4} - 3 z = 9$

Factor $2$ out of $4$ .

$2 \cdot 7 \frac{3}{2 \cdot 2} - 3 z = 9$

Cancel the common factor.

$2 \cdot 7 \frac{3}{2 \cdot 2} - 3 z = 9$

Rewrite the expression.

$7 (\frac{3}{2}) - 3 z = 9$

Combine $7$ and $\frac{3}{2}$ .

$\frac{7 \cdot 3}{2} - 3 z = 9$

Multiply $7$ by $3$ .

$\frac{21}{2} - 3 z = 9$

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

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Subtract $\frac{21}{2}$ from both sides of the equation.

$- 3 z = 9 - \frac{21}{2}$

To write $9$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- 3 z = 9 \cdot \frac{2}{2} - \frac{21}{2}$

Combine $9$ and $\frac{2}{2}$ .

$- 3 z = \frac{9 \cdot 2}{2} - \frac{21}{2}$

Combine the numerators over the common denominator.

$- 3 z = \frac{9 \cdot 2 - 21}{2}$

Simplify the numerator.

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Multiply $9$ by $2$ .

$- 3 z = \frac{18 - 21}{2}$

Subtract $21$ from $18$ .

$- 3 z = \frac{- 3}{2}$

Move the negative in front of the fraction.

$- 3 z = - \frac{3}{2}$

Divide each term by $- 3$ and simplify.

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Divide each term in $- 3 z = - \frac{3}{2}$ by $- 3$ .

$\frac{- 3 z}{- 3} = - \frac{3}{2} \cdot (- \frac{1}{3})$

Cancel the common factor of $- 3$ .

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Cancel the common factor.

$\frac{- 3 z}{- 3} = - \frac{3}{2} \cdot (- \frac{1}{3})$

Divide $z$ by $1$ .

$z = - \frac{3}{2} \cdot (- \frac{1}{3})$

Simplify $- \frac{3}{2} \cdot (- \frac{1}{3})$ .

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Cancel the common factor of $3$ .

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Move the leading negative in $- \frac{3}{2}$ into the numerator.

$z = \frac{- 3}{2} \cdot (- \frac{1}{3})$

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$z = \frac{- 3}{2} \cdot \frac{- 1}{3}$

Factor $3$ out of $- 3$ .

$z = \frac{3 (- 1)}{2} \cdot \frac{- 1}{3}$

Cancel the common factor.

$z = \frac{3 \cdot - 1}{2} \cdot \frac{- 1}{3}$

Rewrite the expression.

$z = \frac{- 1}{2} \cdot - 1$

Combine $\frac{- 1}{2}$ and $- 1$ .

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

Multiply $- 1$ by $- 1$ .

$z = \frac{1}{2}$

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

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

$4 (\frac{3}{4}) + y - 2 (\frac{1}{2}) = 0$

Solve for $y$ .

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Simplify $4 (\frac{3}{4}) + y - 2 (\frac{1}{2})$ .

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Simplify each term.

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Cancel the common factor of $4$ .

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Cancel the common factor.

$4 (\frac{3}{4}) + y - 2 (\frac{1}{2}) = 0$

Rewrite the expression.

$3 + y - 2 (\frac{1}{2}) = 0$

Cancel the common factor of $2$ .

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Factor $2$ out of $- 2$ .

$3 + y + 2 (- 1) \frac{1}{2} = 0$

Cancel the common factor.

$3 + y + 2 \cdot - 1 \frac{1}{2} = 0$

Rewrite the expression.

$3 + y - 1 = 0$

Subtract $1$ from $3$ .

$y + 2 = 0$

Subtract $2$ from both sides of the equation.

$y = - 2$

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

$(\frac{3}{4}, - 2, \frac{1}{2})$

The result can be shown in multiple forms.

Point Form:

$(\frac{3}{4}, - 2, \frac{1}{2})$

Equation Form:

$x = \frac{3}{4}, y = - 2, z = \frac{1}{2}$

Enter YOUR Problem

	$1$	$2$	$x$	$+$	$3$	$y$	$-$	$6$	$z$	$=$	$0$
$+$		$2$	$x$	$-$	$3$	$y$	$+$	$3$	$z$	$=$	$9$
	$1$	$4$	$x$				$-$	$3$	$z$	$=$	$9$

		$-$	$4$	$x$	$+$	$6$	$y$	$-$	$6$	$z$	$=$	$-$	$1$	$8$
$+$	$-$	$1$	$8$	$x$	$-$	$6$	$y$	$+$	$3$	$z$	$=$			$0$
	$-$	$2$	$2$	$x$				$-$	$3$	$z$	$=$	$-$	$1$	$8$

	$-$	$1$	$4$	$x$	$+$	$3$	$z$	$=$		$-$	$9$
$+$	$-$	$2$	$2$	$x$	$-$	$3$	$z$	$=$	$-$	$1$	$8$
	$-$	$3$	$6$	$x$				$=$	$-$	$2$	$7$

	$1$	$2$	$x$	$+$	$3$	$y$	$-$	$6$	$z$	$=$	$0$
$+$		$2$	$x$	$-$	$3$	$y$	$+$	$3$	$z$	$=$	$9$
	$1$	$4$	$x$				$-$	$3$	$z$	$=$	$9$

		$-$	$4$	$x$	$+$	$6$	$y$	$-$	$6$	$z$	$=$	$-$	$1$	$8$
$+$	$-$	$1$	$8$	$x$	$-$	$6$	$y$	$+$	$3$	$z$	$=$			$0$
	$-$	$2$	$2$	$x$				$-$	$3$	$z$	$=$	$-$	$1$	$8$

	$-$	$1$	$4$	$x$	$+$	$3$	$z$	$=$		$-$	$9$
$+$	$-$	$2$	$2$	$x$	$-$	$3$	$z$	$=$	$-$	$1$	$8$
	$-$	$3$	$6$	$x$				$=$	$-$	$2$	$7$

	$1$	$2$	$x$	$+$	$3$	$y$	$-$	$6$	$z$	$=$	$0$
$+$		$2$	$x$	$-$	$3$	$y$	$+$	$3$	$z$	$=$	$9$
	$1$	$4$	$x$				$-$	$3$	$z$	$=$	$9$

		$-$	$4$	$x$	$+$	$6$	$y$	$-$	$6$	$z$	$=$	$-$	$1$	$8$
$+$	$-$	$1$	$8$	$x$	$-$	$6$	$y$	$+$	$3$	$z$	$=$			$0$
	$-$	$2$	$2$	$x$				$-$	$3$	$z$	$=$	$-$	$1$	$8$

	$-$	$1$	$4$	$x$	$+$	$3$	$z$	$=$		$-$	$9$
$+$	$-$	$2$	$2$	$x$	$-$	$3$	$z$	$=$	$-$	$1$	$8$
	$-$	$3$	$6$	$x$				$=$	$-$	$2$	$7$