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

Question

1045 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#