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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 $y$ .

$4 x + y - 2 z = 0$

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

Eliminate $y$ from the system.

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

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

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

Simplify.

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Multiply $2$ by $0$ .

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

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

Simplify the left side.

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Multiply $2$ by $4 x + y - 2 z$ .

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

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

Apply the distributive property.

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

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

Simplify.

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Multiply $4$ by $2$ .

$8 x + 2 y + 2 (- 2 z) = 0$

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

Multiply $- 2$ by $2$ .

$8 x + 2 y - 4 z = 0$

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

$8 x + 2 y - 4 z = 0$

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

$8 x + 2 y - 4 z = 0$

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

$8 x + 2 y - 4 z = 0$

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

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

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

The resultant equation has $y$ eliminated.

$2 x - 3 z = 0$

$2 x - 3 z = 0$

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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Multiply $- 2$ by $9$ .

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

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

Simplify the left side.

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Multiply $- 2$ by $2 x - 3 y + 3 z$ .

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

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

Apply the distributive property.

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

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

Simplify.

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Multiply $2$ by $- 2$ .

$- 4 x - 2 (- 3 y) - 2 (3 z) = - 18$

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

Multiply $- 3$ by $- 2$ .

$- 4 x + 6 y - 2 (3 z) = - 18$

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

Multiply $3$ by $- 2$ .

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

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

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

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

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

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

Multiply $3$ by $0$ .

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

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

Simplify the left side.

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Multiply $3$ by $- 6 x - 2 y + z$ .

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

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

Apply the distributive property.

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

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

Simplify.

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Multiply $- 6$ by $3$ .

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

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

Multiply $- 2$ by $3$ .

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

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

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

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

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

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

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

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

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.

$- 22 x - 3 z = - 18$

$- 22 x - 3 z = - 18$

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

$2 x - 3 z = 0$

$- 22 x - 3 z = - 18$

Eliminate $z$ from the system.

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

$(- 1) \cdot (2 x - 3 z) = (- 1) (0)$

$- 22 x - 3 z = - 18$

Simplify.

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Multiply $- 1$ by $0$ .

$(- 1) \cdot (2 x - 3 z) = 0$

$- 22 x - 3 z = - 18$

Simplify the left side.

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Multiply $- 1$ by $2 x - 3 z$ .

$(- 1) (2 x - 3 z) = 0$

$- 22 x - 3 z = - 18$

Apply the distributive property.

$- 1 (2 x) - 1 (- 3 z) = 0$

$- 22 x - 3 z = - 18$

Multiply.

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Multiply $2$ by $- 1$ .

$- 2 x - 1 (- 3 z) = 0$

$- 22 x - 3 z = - 18$

Multiply $- 3$ by $- 1$ .

$- 2 x + 3 z = 0$

$- 22 x - 3 z = - 18$

$- 2 x + 3 z = 0$

$- 22 x - 3 z = - 18$

$- 2 x + 3 z = 0$

$- 22 x - 3 z = - 18$

$- 2 x + 3 z = 0$

$- 22 x - 3 z = - 18$

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

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

The resultant equation has $z$ eliminated.

$- 24 x = - 18$

Divide each term by $- 24$ and simplify.

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Divide each term in $- 24 x = - 18$ by $- 24$ .

$- \frac{24 x}{- 24} = - \frac{18}{- 24}$

Simplify the left side of the equation by cancelling the common factors.

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Reduce the expression by cancelling the common factors.

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Factor $24$ out of $24 x$ .

$- \frac{24 (x)}{- 24} = - \frac{18}{- 24}$

Move the negative one from the denominator of $\frac{x}{- 1}$ .

$- (- 1 \cdot x) = - \frac{18}{- 24}$

$- (- 1 \cdot x) = - \frac{18}{- 24}$

Rewrite $- 1 \cdot x$ as $- x$ .

$x = - \frac{18}{- 24}$

$x = - \frac{18}{- 24}$

Simplify $- \frac{18}{- 24}$ .

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Reduce the expression by cancelling the common factors.

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Factor $6$ out of $18$ .

$x = - \frac{6 (3)}{- 24}$

Cancel the common factors.

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Factor $6$ out of $- 24$ .

$x = - \frac{6 \cdot 3}{6 \cdot - 4}$

Cancel the common factor.

$x = - \frac{6 \cdot 3}{6 \cdot - 4}$

Rewrite the expression.

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

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

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

Move the negative in front of the fraction.

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

Simplify $- - \frac{3}{4}$ .

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Multiply $- 1$ by $- 1$ .

$x = 1 (\frac{3}{4})$

Multiply $\frac{3}{4}$ by $1$ .

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

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

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

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

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

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.

$2 (\frac{3}{4}) - 3 z = 0$

Solve for $z$ .

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Simplify each term.

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Cancel the common factor of $2$ .

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Write $2$ as a fraction with denominator $1$ .

$\frac{2}{1} \cdot \frac{3}{4} - 3 z = 0$

Factor out the greatest common factor $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Multiply $\frac{1}{1}$ and $\frac{3}{2}$ .

$\frac{3}{2} - 3 z = 0$

$\frac{3}{2} - 3 z = 0$

Subtract $\frac{3}{2}$ from both sides of the equation.

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

Divide each term by $- 3$ and simplify.

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

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

Simplify the left side of the equation by cancelling the common factors.

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Reduce the expression by cancelling the common factors.

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Factor $3$ out of $3 z$ .

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

Move the negative one from the denominator of $\frac{z}{- 1}$ .

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

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

Rewrite $- 1 \cdot z$ as $- z$ .

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

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

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 out the greatest common factor $3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Simplify.

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

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

Multiply $- 1$ by $- 1$ .

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

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

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

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

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

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

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 the left side.

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Simplify each term.

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Cancel the common factor of $4$ .

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Write $4$ as a fraction with denominator $1$ .

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

Factor out the greatest common factor $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Simplify.

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Multiply $\frac{1}{1}$ and $\frac{3}{1}$ .

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

Divide $3$ by $1$ .

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

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

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

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Write $- 2$ as a fraction with denominator $1$ .

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

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

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

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

Divide $- 2$ by $2$ .

$3 + y - 1 = 0$

$3 + y - 1 = 0$

Subtract $1$ from $3$ .

$y + 2 = 0$

$y + 2 = 0$

Subtract $2$ from both sides of the equation.

$y = - 2$

$y = - 2$

$y = - 2$

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:

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

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$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$