How do you solve the system $4c+2d=10$ and $c+3d=10$ by substitution?

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Answer 1 · 2015-05-27T13:31:40

Given
$"[1] " 4c+2d = 10$
$"[2] " c+3d = 10$

[1] can be re-expressed as
$"[3] "d = 5-2c$

then [3] can be substituted into [2]
$"[4] "c+3(5-2c) = 10$

Simplifying

$-5c +15 = 10$
or
$"[5] "c=1$

Substituting [5] back into [2]

$1+3d =10$
or
$"[6] "d=3$

The solution is $(c,d) = (1,4)$

How do you solve the system 4c+2d=104c+2d=10 and c+3d=10c+3d=10 by substitution?