Solve by Addition/Elimination 4b+13c=-24 12b-5c=16

$4 b + 13 c = - 24$ $12 b - 5 c = 16$

Step 1

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

$(- 3) \cdot (4 b + 13 c) = (- 3) (- 24)$

$12 b - 5 c = 16$

Step 2

Simplify.

Simplify the left side.

Simplify $(- 3) \cdot (4 b + 13 c)$ .

Apply the distributive property.

$- 3 (4 b) - 3 (13 c) = (- 3) (- 24)$

$12 b - 5 c = 16$

Multiply.

Multiply $4$ by $- 3$ .

$- 12 b - 3 (13 c) = (- 3) (- 24)$

$12 b - 5 c = 16$

Multiply $13$ by $- 3$ .

$- 12 b - 39 c = (- 3) (- 24)$

$12 b - 5 c = 16$

$- 12 b - 39 c = (- 3) (- 24)$

$12 b - 5 c = 16$

$- 12 b - 39 c = (- 3) (- 24)$

$12 b - 5 c = 16$

$- 12 b - 39 c = (- 3) (- 24)$

$12 b - 5 c = 16$

Simplify the right side.

Multiply $- 3$ by $- 24$ .

$- 12 b - 39 c = 72$

$12 b - 5 c = 16$

$- 12 b - 39 c = 72$

$12 b - 5 c = 16$

$- 12 b - 39 c = 72$

$12 b - 5 c = 16$

Step 3

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

	$-$	$1$	$2$	$b$	$-$	$3$	$9$	$c$	$=$	$7$	$2$
$+$		$1$	$2$	$b$		$-$	$5$	$c$	$=$	$1$	$6$
					$-$	$4$	$4$	$c$	$=$	$8$	$8$

Step 4

Divide each term in $- 44 c = 88$ by $- 44$ and simplify.

Divide each term in $- 44 c = 88$ by $- 44$ .

$\frac{- 44 c}{- 44} = \frac{88}{- 44}$

Simplify the left side.

Cancel the common factor of $- 44$ .

Cancel the common factor.

$\frac{- 44 c}{- 44} = \frac{88}{- 44}$

Divide $c$ by $1$ .

$c = \frac{88}{- 44}$

Simplify the right side.

Divide $88$ by $- 44$ .

$c = - 2$

Step 5

Substitute the value found for $c$ into one of the original equations, then solve for $b$ .

Substitute the value found for $c$ into one of the original equations to solve for $b$ .

$- 12 b - 39 \cdot - 2 = 72$

Multiply $- 39$ by $- 2$ .

$- 12 b + 78 = 72$

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

Subtract $78$ from both sides of the equation.

$- 12 b = 72 - 78$

Subtract $78$ from $72$ .

$- 12 b = - 6$

Divide each term in $- 12 b = - 6$ by $- 12$ and simplify.

Divide each term in $- 12 b = - 6$ by $- 12$ .

$\frac{- 12 b}{- 12} = \frac{- 6}{- 12}$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 b}{- 12} = \frac{- 6}{- 12}$

Divide $b$ by $1$ .

$b = \frac{- 6}{- 12}$

Simplify the right side.

Cancel the common factor of $- 6$ and $- 12$ .

Factor $- 6$ out of $- 6$ .

$b = \frac{- 6 (1)}{- 12}$

Cancel the common factors.

Factor $- 6$ out of $- 12$ .

$b = \frac{- 6 \cdot 1}{- 6 \cdot 2}$

Cancel the common factor.

$b = \frac{- 6 \cdot 1}{- 6 \cdot 2}$

Rewrite the expression.

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

Step 6

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

$c = - 2$

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

Solve by Addition/Elimination 4b+13c=-24 12b-5c=16

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