How do you solve $9c + 2d = 54$ and $c + 2d = 6$ ?

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Answer 1 · 2018-05-21T16:49:11

$c=6, d=0$

$9c+2d=54-----(1)$

$c+2d=6------(2)$

$:.(1)-(2)$

$:.8c=48$

$:.c=cancel 48^6/cancel 8^1$

$:.c=6$

substitute c=6 in (2)

$:.(6)+2d=6$

$:.6+2d=6$

$:.2d=6-6$

$:.2d=0$

$:.d=0$

~~~~~~~~~~~~~

substitute c=6 and=0 in (1)

$:.9(6)+2(0)=54$

$:.54+0=54$

$:.54=54$

How do you solve 9c + 2d = 549c + 2d = 54 and c + 2d = 6 c + 2d = 6?