Precalculus Examples

$x - 2 y - z = 14$ , $2 x - y - 2 z = 10$

Step 1

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

$(- 2) \cdot (x - 2 y - z) = (- 2) (14)$

$2 x - y - 2 z = 10$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

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

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Step 2.1.1.1

Apply the distributive property.

$- 2 x - 2 (- 2 y) - 2 (- z) = (- 2) (14)$

$2 x - y - 2 z = 10$

Step 2.1.1.2

Simplify.

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Step 2.1.1.2.1

Multiply $- 2$ by $- 2$ .

$- 2 x + 4 y - 2 (- z) = (- 2) (14)$

$2 x - y - 2 z = 10$

Step 2.1.1.2.2

Multiply $- 1$ by $- 2$ .

$- 2 x + 4 y + 2 z = (- 2) (14)$

$2 x - y - 2 z = 10$

$- 2 x + 4 y + 2 z = (- 2) (14)$

$2 x - y - 2 z = 10$

$- 2 x + 4 y + 2 z = (- 2) (14)$

$2 x - y - 2 z = 10$

$- 2 x + 4 y + 2 z = (- 2) (14)$

$2 x - y - 2 z = 10$

Step 2.2

Simplify the right side.

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Step 2.2.1

Multiply $- 2$ by $14$ .

$- 2 x + 4 y + 2 z = - 28$

$2 x - y - 2 z = 10$

$- 2 x + 4 y + 2 z = - 28$

$2 x - y - 2 z = 10$

$- 2 x + 4 y + 2 z = - 28$

$2 x - y - 2 z = 10$

Step 3

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

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

Step 4

Divide each term in $3 y = - 18$ by $3$ and simplify.

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Step 4.1

Divide each term in $3 y = - 18$ by $3$ .

$\frac{3 y}{3} = \frac{- 18}{3}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $3$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{- 18}{3}$

Step 4.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 18}{3}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $- 18$ by $3$ .

$y = - 6$

Step 5

Substitute the value found for $y$ into one of the original equations, then solve for $x$ .

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Step 5.1

Substitute the value found for $y$ into one of the original equations to solve for $x$ .

$- 2 x + 4 (- 6) + 2 z = - 28$

Step 5.2

Multiply $4$ by $- 6$ .

$- 2 x - 24 + 2 z = - 28$

Step 5.3

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

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Step 5.3.1

Add $24$ to both sides of the equation.

$- 2 x + 2 z = - 28 + 24$

Step 5.3.2

Subtract $2 z$ from both sides of the equation.

$- 2 x = - 28 + 24 - 2 z$

Step 5.3.3

Add $- 28$ and $24$ .

$- 2 x = - 4 - 2 z$

Step 5.4

Divide each term in $- 2 x = - 4 - 2 z$ by $- 2$ and simplify.

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Step 5.4.1

Divide each term in $- 2 x = - 4 - 2 z$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 4}{- 2} + \frac{- 2 z}{- 2}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $- 2$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 4}{- 2} + \frac{- 2 z}{- 2}$

Step 5.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 4}{- 2} + \frac{- 2 z}{- 2}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Simplify each term.

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Step 5.4.3.1.1

Divide $- 4$ by $- 2$ .

$x = 2 + \frac{- 2 z}{- 2}$

Step 5.4.3.1.2

Cancel the common factor of $- 2$ .

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Step 5.4.3.1.2.1

Cancel the common factor.

$x = 2 + \frac{- 2 z}{- 2}$

Step 5.4.3.1.2.2

Divide $z$ by $1$ .

$x = 2 + z$

Step 6

This is the final solution to the independent system of equations.

$y = - 6$

$x = 2 + z$

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

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

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