How do you solve the system of equations #-2x+4y=7# and ##?

1 Answer
Nov 5, 2017

Suppose the second equation was : #4x-y=0#
then
#color(white)("XXX")(x,y)=(1/2,2)#

Explanation:

The only way to clear this question (after a year with no correction) seems to be to add the missing component. The additional equation #4x-y=0# was completely arbitrary.

[1]#color(white)("XXX")-2x+4y=7#
[2]#color(white)("XXX")4x-y=0#

Multiplying [1] by #2#
[3]#color(white)("XXX")-4x+8y=14#

Adding [2] and [3]
[4]#color(white)("XXX")7y=14#

Dividing both sides of [4] by #7#
[5]#color(white)("XXX")y=2#

Substituting #2# for #y# in [2]
[6]#color(white)("XXX")4x-2=0#

[7]#color(white)("XXX")4x=2#

[8]#color(white)("XXX")x=1/2#