How do you solve the system #x/3 + 3y/4 = -3# and #x/3 - y/4 = 5#?

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Answer 1 · 2016-06-29T17:07:03

#x=9# and #y=-8#

The equations are:

#x/3+(3y)/4=-3# ...................(A)

#x/3-y/4=5# ...................(B)

Now subtracting (B) from (A)

#(3y)/4+y/4=-3-5=-8# or #y=-8#

Putting this in (B)

#x/3-(-8)/4=5# or #x/3+2=5# or #x/3=3# or #x=9#