Precalculus Examples

$4 a + 7 b = - 24$ , $8 a - 3 c = - 7$ , $6 b - 3 c = - 39$

Step 1

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

$4 a + 7 b = - 24$

$8 a - 3 c = - 7$

Step 2

Eliminate $a$ from the system.

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Step 2.1

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

$(- 2) \cdot (4 a + 7 b) = (- 2) (- 24)$

$8 a - 3 c = - 7$

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 $(- 2) \cdot (4 a + 7 b)$ .

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Step 2.2.1.1.1

Apply the distributive property.

$- 2 (4 a) - 2 (7 b) = (- 2) (- 24)$

$8 a - 3 c = - 7$

Step 2.2.1.1.2

Multiply.

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Step 2.2.1.1.2.1

Multiply $4$ by $- 2$ .

$- 8 a - 2 (7 b) = (- 2) (- 24)$

$8 a - 3 c = - 7$

Step 2.2.1.1.2.2

Multiply $7$ by $- 2$ .

$- 8 a - 14 b = (- 2) (- 24)$

$8 a - 3 c = - 7$

$- 8 a - 14 b = (- 2) (- 24)$

$8 a - 3 c = - 7$

$- 8 a - 14 b = (- 2) (- 24)$

$8 a - 3 c = - 7$

$- 8 a - 14 b = (- 2) (- 24)$

$8 a - 3 c = - 7$

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

Multiply $- 2$ by $- 24$ .

$- 8 a - 14 b = 48$

$8 a - 3 c = - 7$

$- 8 a - 14 b = 48$

$8 a - 3 c = - 7$

$- 8 a - 14 b = 48$

$8 a - 3 c = - 7$

Step 2.3

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

			$-$	$8$	$a$	$-$	$1$	$4$	$b$	$=$	$4$	$8$
		$+$		$8$	$a$					$=$	$-$	$7$
$-$	$3$	$c$				$-$	$1$	$4$	$b$	$=$	$4$	$1$

Step 2.4

The resultant equation has $a$ eliminated.

$- 14 b - 3 c = 41$

Step 3

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

$- 14 b - 3 c = 41$

$6 b - 3 c = - 39$

Step 4

Eliminate $c$ from the system.

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Step 4.1

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

$(- 1) \cdot (- 14 b - 3 c) = (- 1) (41)$

$6 b - 3 c = - 39$

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 $(- 1) \cdot (- 14 b - 3 c)$ .

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Step 4.2.1.1.1

Apply the distributive property.

$- 1 (- 14 b) - 1 (- 3 c) = (- 1) (41)$

$6 b - 3 c = - 39$

Step 4.2.1.1.2

Multiply.

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Step 4.2.1.1.2.1

Multiply $- 14$ by $- 1$ .

$14 b - 1 (- 3 c) = (- 1) (41)$

$6 b - 3 c = - 39$

Step 4.2.1.1.2.2

Multiply $- 3$ by $- 1$ .

$14 b + 3 c = (- 1) (41)$

$6 b - 3 c = - 39$

$14 b + 3 c = (- 1) (41)$

$6 b - 3 c = - 39$

$14 b + 3 c = (- 1) (41)$

$6 b - 3 c = - 39$

$14 b + 3 c = (- 1) (41)$

$6 b - 3 c = - 39$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Multiply $- 1$ by $41$ .

$14 b + 3 c = - 41$

$6 b - 3 c = - 39$

$14 b + 3 c = - 41$

$6 b - 3 c = - 39$

$14 b + 3 c = - 41$

$6 b - 3 c = - 39$

Step 4.3

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

	$1$	$4$	$b$	$+$	$3$	$c$	$=$	$-$	$4$	$1$
$+$		$6$	$b$	$-$	$3$	$c$	$=$	$-$	$3$	$9$
	$2$	$0$	$b$				$=$	$-$	$8$	$0$

Step 4.4

The resultant equation has $c$ eliminated.

$20 b = - 80$

Step 4.5

Divide each term in $20 b = - 80$ by $20$ and simplify.

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Step 4.5.1

Divide each term in $20 b = - 80$ by $20$ .

$\frac{20 b}{20} = \frac{- 80}{20}$

Step 4.5.2

Simplify the left side.

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Step 4.5.2.1

Cancel the common factor of $20$ .

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Step 4.5.2.1.1

Cancel the common factor.

$\frac{20 b}{20} = \frac{- 80}{20}$

Step 4.5.2.1.2

Divide $b$ by $1$ .

$b = \frac{- 80}{20}$

Step 4.5.3

Simplify the right side.

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Step 4.5.3.1

Divide $- 80$ by $20$ .

$b = - 4$

Step 5

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

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Step 5.1

Substitute the value of $b$ into an equation with $a$ eliminated already.

$- 14 \cdot - 4 - 3 c = 41$

Step 5.2

Solve for $c$ .

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Step 5.2.1

Multiply $- 14$ by $- 4$ .

$56 - 3 c = 41$

Step 5.2.2

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

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Step 5.2.2.1

Subtract $56$ from both sides of the equation.

$- 3 c = 41 - 56$

Step 5.2.2.2

Subtract $56$ from $41$ .

$- 3 c = - 15$

Step 5.2.3

Divide each term in $- 3 c = - 15$ by $- 3$ and simplify.

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Step 5.2.3.1

Divide each term in $- 3 c = - 15$ by $- 3$ .

$\frac{- 3 c}{- 3} = \frac{- 15}{- 3}$

Step 5.2.3.2

Simplify the left side.

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Step 5.2.3.2.1

Cancel the common factor of $- 3$ .

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Step 5.2.3.2.1.1

Cancel the common factor.

$\frac{- 3 c}{- 3} = \frac{- 15}{- 3}$

Step 5.2.3.2.1.2

Divide $c$ by $1$ .

$c = \frac{- 15}{- 3}$

Step 5.2.3.3

Simplify the right side.

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Step 5.2.3.3.1

Divide $- 15$ by $- 3$ .

$c = 5$

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.

$4 a + 7 (- 4) = - 24$

Step 6.2

Solve for $a$ .

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Step 6.2.1

Multiply $7$ by $- 4$ .

$4 a - 28 = - 24$

Step 6.2.2

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

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Step 6.2.2.1

Add $28$ to both sides of the equation.

$4 a = - 24 + 28$

Step 6.2.2.2

Add $- 24$ and $28$ .

$4 a = 4$

Step 6.2.3

Divide each term in $4 a = 4$ by $4$ and simplify.

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Step 6.2.3.1

Divide each term in $4 a = 4$ by $4$ .

$\frac{4 a}{4} = \frac{4}{4}$

Step 6.2.3.2

Simplify the left side.

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Step 6.2.3.2.1

Cancel the common factor of $4$ .

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Step 6.2.3.2.1.1

Cancel the common factor.

$\frac{4 a}{4} = \frac{4}{4}$

Step 6.2.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{4}{4}$

Step 6.2.3.3

Simplify the right side.

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Step 6.2.3.3.1

Divide $4$ by $4$ .

$a = 1$

Step 7

This is the final solution to the system of equations.

$b = - 4$

$c = 5$

$a = 1$

			$-$	$8$	$a$	$-$	$1$	$4$	$b$	$=$	$4$	$8$
		$+$		$8$	$a$					$=$	$-$	$7$
$-$	$3$	$c$				$-$	$1$	$4$	$b$	$=$	$4$	$1$

	$1$	$4$	$b$	$+$	$3$	$c$	$=$	$-$	$4$	$1$
$+$		$6$	$b$	$-$	$3$	$c$	$=$	$-$	$3$	$9$
	$2$	$0$	$b$				$=$	$-$	$8$	$0$

			$-$	$8$	$a$	$-$	$1$	$4$	$b$	$=$	$4$	$8$
		$+$		$8$	$a$					$=$	$-$	$7$
$-$	$3$	$c$				$-$	$1$	$4$	$b$	$=$	$4$	$1$

	$1$	$4$	$b$	$+$	$3$	$c$	$=$	$-$	$4$	$1$
$+$		$6$	$b$	$-$	$3$	$c$	$=$	$-$	$3$	$9$
	$2$	$0$	$b$				$=$	$-$	$8$	$0$

			$-$	$8$	$a$	$-$	$1$	$4$	$b$	$=$	$4$	$8$
		$+$		$8$	$a$					$=$	$-$	$7$
$-$	$3$	$c$				$-$	$1$	$4$	$b$	$=$	$4$	$1$

	$1$	$4$	$b$	$+$	$3$	$c$	$=$	$-$	$4$	$1$
$+$		$6$	$b$	$-$	$3$	$c$	$=$	$-$	$3$	$9$
	$2$	$0$	$b$				$=$	$-$	$8$	$0$