Precalculus Examples

$8 x - z = 4$ , $y + z = 5$ , $11 x + y = 15$

Step 1

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

$8 x - z = 4$

$y + z = 5$

Step 2

Eliminate $z$ from the system.

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Step 2.1

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

	$8$	$x$	$-$	$z$	$=$	$4$
$+$			$+$	$z$	$=$	$5$
$y$	$8$	$x$	$+$		$=$	$9$

Step 2.2

The resultant equation has $z$ eliminated.

$8 x + y = 9$

Step 3

Take the resultant equation and the third original equation and eliminate another variable. In this case, eliminate $y$ .

$8 x + y = 9$

$11 x + y = 15$

Step 4

Eliminate $y$ from the system.

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Step 4.1

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

$8 x + y = 9$

$(- 1) \cdot (11 x + y) = (- 1) (15)$

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 $(- 1) \cdot (11 x + y)$ .

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Step 4.2.1.1.1

Apply the distributive property.

$8 x + y = 9$

$- 1 (11 x) - 1 y = (- 1) (15)$

Step 4.2.1.1.2

Simplify the expression.

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Step 4.2.1.1.2.1

Multiply $11$ by $- 1$ .

$8 x + y = 9$

$- 11 x - 1 y = (- 1) (15)$

Step 4.2.1.1.2.2

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

$8 x + y = 9$

$- 11 x - y = (- 1) (15)$

$8 x + y = 9$

$- 11 x - y = (- 1) (15)$

$8 x + y = 9$

$- 11 x - y = (- 1) (15)$

$8 x + y = 9$

$- 11 x - y = (- 1) (15)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Multiply $- 1$ by $15$ .

$8 x + y = 9$

$- 11 x - y = - 15$

$8 x + y = 9$

$- 11 x - y = - 15$

$8 x + y = 9$

$- 11 x - y = - 15$

Step 4.3

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

			$8$	$x$	$+$	$y$	$=$			$9$
$+$	$-$	$1$	$1$	$x$	$-$	$y$	$=$	$-$	$1$	$5$
		$-$	$3$	$x$			$=$		$-$	$6$

Step 4.4

The resultant equation has $y$ eliminated.

$- 3 x = - 6$

Step 4.5

Divide each term in $- 3 x = - 6$ by $- 3$ and simplify.

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Step 4.5.1

Divide each term in $- 3 x = - 6$ by $- 3$ .

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

Step 4.5.2

Simplify the left side.

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Step 4.5.2.1

Cancel the common factor of $- 3$ .

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Step 4.5.2.1.1

Cancel the common factor.

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

Step 4.5.2.1.2

Divide $x$ by $1$ .

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

Step 4.5.3

Simplify the right side.

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Step 4.5.3.1

Divide $- 6$ by $- 3$ .

$x = 2$

Step 5

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

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Step 5.1

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

$8 (2) + y = 9$

Step 5.2

Solve for $y$ .

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Step 5.2.1

Multiply $8$ by $2$ .

$16 + y = 9$

Step 5.2.2

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

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Step 5.2.2.1

Subtract $16$ from both sides of the equation.

$y = 9 - 16$

Step 5.2.2.2

Subtract $16$ from $9$ .

$y = - 7$

Step 6

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

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Step 6.1

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

$8 (2) - z = 4$

Step 6.2

Solve for $z$ .

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Step 6.2.1

Multiply $8$ by $2$ .

$16 - z = 4$

Step 6.2.2

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

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Step 6.2.2.1

Subtract $16$ from both sides of the equation.

$- z = 4 - 16$

Step 6.2.2.2

Subtract $16$ from $4$ .

$- z = - 12$

Step 6.2.3

Divide each term in $- z = - 12$ by $- 1$ and simplify.

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Step 6.2.3.1

Divide each term in $- z = - 12$ by $- 1$ .

$\frac{- z}{- 1} = \frac{- 12}{- 1}$

Step 6.2.3.2

Simplify the left side.

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Step 6.2.3.2.1

Dividing two negative values results in a positive value.

$\frac{z}{1} = \frac{- 12}{- 1}$

Step 6.2.3.2.2

Divide $z$ by $1$ .

$z = \frac{- 12}{- 1}$

Step 6.2.3.3

Simplify the right side.

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Step 6.2.3.3.1

Divide $- 12$ by $- 1$ .

$z = 12$

Step 7

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

$(2, - 7, 12)$

Step 8

The result can be shown in multiple forms.

Point Form:

$(2, - 7, 12)$

Equation Form:

$x = 2, y = - 7, z = 12$

			$8$	$x$	$+$	$y$	$=$			$9$
$+$	$-$	$1$	$1$	$x$	$-$	$y$	$=$	$-$	$1$	$5$
		$-$	$3$	$x$			$=$		$-$	$6$

			$8$	$x$	$+$	$y$	$=$			$9$
$+$	$-$	$1$	$1$	$x$	$-$	$y$	$=$	$-$	$1$	$5$
		$-$	$3$	$x$			$=$		$-$	$6$

			$8$	$x$	$+$	$y$	$=$			$9$
$+$	$-$	$1$	$1$	$x$	$-$	$y$	$=$	$-$	$1$	$5$
		$-$	$3$	$x$			$=$		$-$	$6$