Question #02b23

1 Answer
Alan P.
Mar 29, 2017

#(x,y)=(-4/3,2/3)#

Explanation:

I've no idea what the leading #(-3,1)# means.

Since this was asked under the heading "Systems Using Substitution", I will demonstrate how to solve the given two equations (using substitution.

Given
[1]#color(white)("XXX")x-y=-2#
[2]#color(white)("XXX")x+5y=2#

Rewriting [1] as a definition in terms of #x#
[3]#color(white)("XXX")x=y-2#

We can now substitute #y-2# for #x# in equation [2]
[4]#color(white)("XXX")(y-2) +5y=2#

#rarr#[5]#color(white)("XXX")6y=4#

#rarr#[6]#color(white)("XXX")y=2/3#

Now substitute #2/3# for #y# in [1]
[7]#color(white)("XXX")x-2/3=-2#

#rarr#[8]#color(white)("XXX")x=-6/3+2/3=-4/3#

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