How do you solve #2x - 4y = 32# and #8x - 8y = 96 #?

1 Answer
Feb 6, 2018

Notice that we're gonna solve for #x# first because if we solve for #x# in the first equation... we can factorize and get #x# without doing any kind of multiplication or something

Explanation:

We're gonna solve for #x# first

First equation
#2x-4y=32#
#2x=32+4y#
#x=(32+4y)/2#
#x=32/2+(4y)/2#
#x=16+2y#

Second equation
#8x-8y =96#
Factorize
#8(x-y)=96#
Transfer the 8 and put value of #x#
#16+2y-y=96/8#
#16+y=12#
#y=12-16#
#color(red)(y=-4)#
Get the value of #x#
#x=16+2y#
#x=16+2xx-4#
#x=16+(-8)#
#color(red)(x=8)#