Given
[1]#color(white)("XXX")2x+4y=6#
[2]#color(white)("XXX")4x-3y=-10#
Method 1: By elimination
We note that if we multiply equation [1] by #2# the coefficient of #x# will be the same as that of equation [2]
[3]#color(white)("XXX")4x+8y=12#
We can now eliminate the #x# term by subtracting equation [2] from equation [3]
#color(white)("XXX.."[x])4x+8y=color(white)("x..")12#
#color(white)("XXX")-(ul(4x-3y=-10))#
[4]#color(white)("XXXXXx..")11y=color(white)("x..")22#
After dividing both sides by #11#
[5]#color(white)("XXX")y=2#
Then substituting #2# for #y# in [1]
[6]#color(white)("XXX")2x+4 * 2 =6#
which simplifies as
[7]#color(white)("XXX")2x=-2#
or
[8]#color(white)("XXX")x=-1#
Method 2: By substitution
We note that we can divide both sides of [1] by #2# and then re-arrange the terms to get
[9]#color(white)("XXX")x=3-2y#
Now we can substitute #(3-2y)# for #x# in [2]
[10]#color(white)("XXX")4 * (3-2y)-3y=-10#
Simplifying [10]
[11]#color(white)("XXX")12-8y-3y=-10#
then
[12]#color(white)("XXX")-11y=-22#
and finally (after dividing both sides by #2#
[13]#color(white)("XXX")y=2#
Substituting #(2)# for #y# back in [1] (or [9]) gives
[14}#color(white)("XXX")x=-1#
