Solve by Addition/Elimination 2c-15d+7e=-30 , -c+8d-4e=16 , -c+3d+e=6

$2 c - 15 d + 7 e = - 30$ , $- c + 8 d - 4 e = 16$ , $- c + 3 d + e = 6$

To solve a system of $2$ variables, only $2$ equations are required. Choose the first two equations containing the variables in the system.

$2 c - 15 d + 7 e = - 30$

$- c + 8 d - 4 e = 16$

Subtract $7 e$ from both sides of the equation.

$2 c - 15 d = - 30 - 7 e$

$- c + 8 d - 4 e = 16$

Add $4 e$ to both sides of the equation.

$- c + 8 d = 16 + 4 e$

$2 c - 15 d = - 30 - 7 e$

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

$2 c - 15 d = - 30 - 7 e$

$(2) \cdot (- c + 8 d) = (2) (16 + 4 e)$

Simplify.

Simplify the left side.

Simplify $(2) \cdot (- c + 8 d)$ .

Apply the distributive property.

$2 c - 15 d = - 30 - 7 e$

$2 (- c) + 2 (8 d) = (2) (16 + 4 e)$

Multiply.

Multiply $- 1$ by $2$ .

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 2 (8 d) = (2) (16 + 4 e)$

Multiply $8$ by $2$ .

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 16 d = (2) (16 + 4 e)$

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 16 d = (2) (16 + 4 e)$

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 16 d = (2) (16 + 4 e)$

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 16 d = (2) (16 + 4 e)$

Simplify the right side.

Simplify $(2) (16 + 4 e)$ .

Apply the distributive property.

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 16 d = 2 \cdot 16 + 2 (4 e)$

Multiply.

Multiply $2$ by $16$ .

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 16 d = 32 + 2 (4 e)$

Multiply $4$ by $2$ .

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 16 d = 32 + 8 e$

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 16 d = 32 + 8 e$

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 16 d = 32 + 8 e$

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 16 d = 32 + 8 e$

$2 c - 15 d = - 30 - 7 e$

$- 2 c + 16 d = 32 + 8 e$

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

		$2$	$c$	$-$	$1$	$5$	$d$	$=$	$-$	$3$	$0$	$-$	$7$	$e$
$+$	$-$	$2$	$c$	$+$	$1$	$6$	$d$	$=$		$3$	$2$	$+$	$8$	$e$
							$d$	$=$			$2$	$+$		$e$

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

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

$2 c - 15 (2 + e) = - 30 - 7 e$

Simplify each term.

Apply the distributive property.

$2 c - 15 \cdot 2 - 15 e = - 30 - 7 e$

Multiply $- 15$ by $2$ .

$2 c - 30 - 15 e = - 30 - 7 e$

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

Add $30$ to both sides of the equation.

$2 c - 15 e = - 30 - 7 e + 30$

Add $15 e$ to both sides of the equation.

$2 c = - 30 - 7 e + 30 + 15 e$

Add $- 30$ and $30$ .

$2 c = 0 - 7 e + 15 e$

Subtract $7 e$ from $0$ .

$2 c = - 7 e + 15 e$

Add $- 7 e$ and $15 e$ .

$2 c = 8 e$

Divide each term in $2 c = 8 e$ by $2$ and simplify.

Divide each term in $2 c = 8 e$ by $2$ .

$\frac{2 c}{2} = \frac{8 e}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 c}{2} = \frac{8 e}{2}$

Divide $c$ by $1$ .

$c = \frac{8 e}{2}$

Simplify the right side.

Cancel the common factor of $8$ and $2$ .

Factor $2$ out of $8 e$ .

$c = \frac{2 (4 e)}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$c = \frac{2 (4 e)}{2 (1)}$

Cancel the common factor.

$c = \frac{2 (4 e)}{2 \cdot 1}$

Rewrite the expression.

$c = \frac{4 e}{1}$

Divide $4 e$ by $1$ .

$c = 4 e$

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

$d = 2 + e$

$c = 4 e$

Solve by Addition/Elimination 2c-15d+7e=-30 , -c+8d-4e=16 , -c+3d+e=6

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