How do you solve the system of equations #4x - 3y = - 38# and #2x + 4y = - 8#?

1 Answer
Jan 12, 2018

#(x,y)=(-8,2)#

Explanation:

Given
[1]#color(white)("XXX")4x-3y=-38#
[2]#color(white)("XXX")2x+4y=-8#

Method 1: Elimination
If we multiply [2] by 2 the#x# term will have the same coefficient as that in [1].;
so after multiplying [2] by 2
[3]#color(white)("XXX")4x+8y=-16#

subtract [3] from [1] to eliminate the #x# term
[4]#color(white)("XXX")-11y=-22#

then divide both sides by #(-11)# to get
[5]#color(white)("XXX")y=2#

Now we can substitute #2# for #y# in one of the original equations (I will use [1])
[6]#color(white)("XXX")2x+4 xx 2 =-8#

subtracting #4xx2=8# from both sides
[7]#color(white)("XXX")2x=-16#

then dividing both sides by #2#
[8]#color(white)("XXX")x=-8#

giving the solution pair #(x,y)=(-8,2)#

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Method 2: Substitution
[2] can be rewritten to isolate the #x# term on one side and everything else on the other:
[9]#color(white)("XXX")2x=-8-4y#

then after dividing both sides by #2#
[10]#color(white)("XXX")x=-4-2y#

we can now substitute #(-4-2y)# for #x# in [1]
[11]#color(white)("XXX")4 xx (-4-2y) -3y=-38#

simplifying
[12]#color(white)("XXX")-16-11y=-38#

adding #16# to both sides
[13]#color(white)("XXX")-11y= -22#

dividing both sides by #(-11)#
[14]#color(white)("XXX")y=2#

Now, substitute #2# for #y# in one of the original equations (I will use [2] this time)
[15]#color(white)("XXX")2x+4xx2=-8#

subtract #4xx2=8# from both sides
[16]#color(white)("XXX")2x=-16#

finally divide both sides by #(-2)#
[17]#color(white)("XXX")x=-8#

again, giving the solution pair #x=-8, y=2#