What is the solution to the system of equations #3x+y=16# and #2x+2y = 6#?

1 Answer
Jun 19, 2018

#x=13/2# and #y=-7/2#

Explanation:

Given
[1]#color(white)("XXX")3x+y=16#
[2]#color(white)("XXX")2x+2y=6#

We will solve this by "elimination"; that is we will attempt to combine the given equations in some way so that we end up with an equation with only one variable (we "eliminate" the other variable).

Looking at the given equations we can see that simply adding or subtracting one from the other will not eliminate either variable;

however, if we first multiply equation [1] by #2# the #y# term will become #2y# and by subtracting equation [2], the #y# term will be eliminated.

[3]=[1]#xx2color(white)("XXX")6x+2y=32#
[2]#color(white)("XXXXxX")-(ul(2x+2y=color(white)("x")6))#
[4]#color(white)("XXXXXxXX-")4xcolor(white)("xxxx")=26#

No we can divide both sides of equation [4] by #4# to get a simple value for #x#

[5]=[4]#div4color(white)("XXX")x=13/2#

Now we can use this value of #x# back in one of the original equations to determine the value of #y#.

For example, substituting #13/2# for #x# in [2]
[6]: [2] with #x=13/2color(white)("XXX")2 * (13/2) +2y =6#

#color(white)("XXXXXXXXXXXXXXXXXXXX")rArr 2y=6-13#

#color(white)("XXXXXXXXXXXXXXXXXXXX")rArr y=-7/2#

[Note: you really should check this result: #x=13/2, y=-7/2# back in [1] to verify the result.