# How do you solve this system of equations: c+ d = 22 and 4c + 2d = 56?

Dec 7, 2017

$c = 6 \mathmr{and} d = 16$

#### Explanation:

$c + d = 22$-------- let this be equation (1) and
$4 c + 2 d = 56$----------let this be equation (2)

(1)$\implies c = 22 - d$

Substitute this value of c in equation(2),

(2)$\implies 4 \left(22 - d\right) + 2 d = 56$

$\implies 88 - 4 d + 2 d = 56$

$\implies - 2 d = 56 - 88 = - 32$

$\implies d = - \frac{32}{-} 2$

$\implies d = 16$

Now substitute this value of $d$ in (1), to find $c$:

$\implies c + 16 = 22$

$c = 22 - 16 = 6$

Therefore $c = 6 \mathmr{and} d = 16$