How do you solve the following system: #2x-y/2=4, 5x+2y=12 #?

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Answer 1 · 2016-02-03T21:25:04

#x = 28/13 , y = 8/13#

This is done by the process of elimination but it is not the only method.

Begin by multiplying the first equation by 4 to get the coefficients in front of the #y# equal. Multiplying the first equation by 4 gives:

#8x -2y =16#

Now we can add both equations together

# (5x +2y) + (8x-2y) = (16)+(12)#

So we get:

#13x = 28 -> x = 28/13#

Now replace #x# with this value in either of the equations and solve for #y#. So using the first equation: # 2x-y/2 =4#.

#-> 2(28/13) - y/2 =4 #
#56/13 - y/2 = 4#

Rearrange and you should arrive at:

#y = 2(56/13 - 4) = 2(56-52)/13=8/13#
#therefore x = 28/13, y = 8/13#