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

Question

1273 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-07T03:58:16

$c= 6 and d=16$

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

(1) $=> c= 22-d$

Substitute this value of c in equation(2),

(2) $=> 4(22-d) +2d = 56$

$=> 88 -4d +2d =56$

$=> -2d = 56-88 = -32$

$=> d= -32/-2$

$=> d= 16$

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

$=>c +16 =22$

$c= 22-16 = 6$

Therefore $c= 6 and d=16$

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