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#