Notice that although the signs are opposite we have #2y# in each of the equations. So if we add them we eliminate the #y# term and are able to solve for #x#. Then by substituting for #x# we can solve for #y#.
#color(blue)("Solving for "x)#
#3x+2y=18" "..................Equation(1)#
#ul(7x-2y=32)" ".................Equation(2)#
#10x+0=50" ".............Eqn(1)+Eqn(2)#
#10x=50#
Divide both sides by 10
#x=5#
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#color(blue)("Solving for "y)#
I choose Eqn(1) because the #y# term is positive.
Substitute for #x# in Eqn(1) giving:
#color(green)(3color(red)(x)+2y=18 color(white)("ddd")->color(white)("ddd")3(color(red)(5))+2y=18)#
#color(green)(color(white)("ddddddd.dddddd") ->color(white)("ddd.d")15+2y=18 )#
Subtract #color(red)(15)# from both sides
#color(green)(15+2y=18 color(white)("ddd")->color(white)("ddd")ubrace(15color(red)(-15)) +2y=18color(red)(-15) )#
#color(green)(color(white)("ddddddddddddd")->color(white)("ddd.dd") 0color(white)("dd")+2y=3)#
Divide both sides by #color(red)(2)#
#color(green)(2y=3 color(white)("dddddddd")->color(white)("dddddddddd")2/color(red)(2)y=3/color(red)(2))#
But #2/2=1# giving
#y=3/2#
