How do you solve the system of equations #10x + 8y = 2# and #- 2x - 4y = 6#?

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Answer 1 · 2018-06-22T18:16:11

#x=7/3 and y=-8/3#

From first equation: #y=(2-10x)/8=(cancel2(1-5x))/cancel8=(1-5x)/4#

Substitute to second equation: #-2x-cancel(4)(1-5x)/cancel(4)=6#

So: #-2x-1+5x= 6 -> x=7/3#

Then: #y=(2-10(7/3))/8=(2-70/3)*1/8=-cancel64/3*1/cancel8=-8/3#

Answer 2 · 2018-06-22T18:19:16

#x=7/3,y=-8/3#

Multiplying the second equation by # 5# and adding to the first we get

#-6y=16#
so

#y=-8/3#
so we get

#-x+16/3=3#

#x=7/3#