How do you solve the system #1/4x+y=3/4# and #3/2x-y=-5/2#?

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Answer 1 · 2015-05-17T12:39:48

From the first equation we can derive a formula for #y# by subtracting #1/4x# from both sides to get

#y = 3/4 - 1/4x#

Then substitute that for #y# in the second equation:

#-5/2 = 3/2x-y#

#= 3/2x-(3/4-1/4x)#

These fractions are a pain, so let's eliminate them by multiplying through by 4 to get:

#-10=6x-(3-x) = 7x-3#

Now add 3 to both sides to get

#7x = -7#

Divide both sides by #7# to get

#x = -1#

Then #y = 3/4 - 1/4x = 3/4 + 1/4 = 1#