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Answer 1 · 2018-02-06T15:05:09

#y=4#

#x=2#

If #4x-2y=0# and #4x+y=12#, then:

#4x-4x+y-(-2y)=12-0#

#=> 0x+y+2y=12 #

#=> 3y=12#

#=> y=4#

Since

#4x+y=12#

we get that

#4x+4=12#

#=> 4x=12-4#

#=> 4x=8#

#=> x=2#

So #y=4# and #x=2#.

Answer 2 · 2018-02-06T15:09:38

#x=2 and y=4#

Notice that the #x# terms are exact;y the same in both equations.

If you subtract the equations, the #x# terms will be eliminated and you will be able to solve for #y#

#" "4x +y =12" "rarr A#
#" "4x -2y =0" "rarr B#

#A-B:" " 0x+3y =12" "rarrC#

#color(white)(wwwwwwwwwww)y = 4#

Once you know the value for #y# you can use it to find #x#

Substitute #4 " for " y " in " A#

#" "4x +y =12" "rarr A#
#" "4x +(4) =12#

#" "4x = 8#
#" "x =2#

Check in #B#

#" "4x -2y =0" "rarr B#
#" "4(2) -2(4) =0#
#" "8-8=0#