Solve system of equations:
#color(red)(4x-2y=-2# and #color(blue)(x+2y=-13#
We need to solve one equation for either #x# or #y#. Then substitute that result into the other equation, replacing either #x# or #y#. Once we have a numerical value for #x# or #y#, substitute it into the other equation to solve for #x# or #y#.
I'm going to start with #color(blue)(x+2y=-13# and solve for #x#.
Subtract #2y# from both sides.
#x=-2y-13#
Now substitute this value for #x# into the other equation: #color(red)(4x-2y=-2#.
#4(-2y-13)-2y=-2#
Expand.
#-8y-52-2y=-2#
Add #52# to both sides.
#-8y-2y=-2+52#
Simplify.
#-10y=50#
Divide both sides by #-10#.
#y=-50/10#
Reduce the fraction.
#y=-5#
Now substitute the value for #y# into the previous equation: #color(blue)(x+2y=-13#.
#x+(2)(-5)=-13#
Simplify.
#x-10=-13#
Add #10# to both sides.
#x+-13+10#
Simplify.
#x=-3#
