# How do you solve 2c - d = - 2 and 4c + d = 20?

Nov 30, 2015

$\left(c , d\right) = \left(3 , 8\right)$

#### Explanation:

Given:
$\left[1\right] \textcolor{w h i t e}{\text{XXX}} 2 c - d = - 2$
$\left[2\right] \textcolor{w h i t e}{\text{XXX}} 4 c + d = 20$

{: (,color(white)("XXX"),2c,-d,=,-2), (+,,4c,+d,=, 20), (,,"-------","-----","----","-----"), ([3],,6c,,=,18) :}

Divide [3] by $6$
$\left[4\right] \textcolor{w h i t e}{\text{XXXXXX}} c = 3$

Substitute $3$ (from [4]) for $c$ in [1]
$\left[5\right] \textcolor{w h i t e}{\text{XXX}} 2 \cdot \left(3\right) - d = - 2$
Simplify
$\left[6\right] \textcolor{w h i t e}{\text{XXX}} 6 - d = - 2$
Subtract $6$ from both sides
$\left[7\right] \textcolor{w h i t e}{\text{XXX}} - d = - 8$
Multiply both sides by $\left(- 1\right)$
$\left[8\right] \textcolor{w h i t e}{\text{XXX}} d = 8$