Notice that both equation have #5y#. So if we subtract one from the other we end up with just 1 unknown. This is solvable.
#8x+5y=+21" "......................Equation(1)#
#ul(3x+5y=-14)" "......................Equation(2)#
#5x+0y=+35larrcolor(white)("d") Eqn(1)-Eqn(2)#
#5x=35#
Divide both sides by 5
#5/5 x=35/5#
#x=7 larr" checked"#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To solve for #y# substitute 7 for #x# in equation(1)
#color(green)(8color(red)(x)+5y=21 color(white)("dddd")->color(white)("dddd") 8(color(red)(7))+5y=21)#
#color(green)(color(white)("dddddddddddddd")->color(white)("dddddd") 56+5y=21)#
Subtract 56 from both sides
#color(green)(color(white)("dddddddddddddd")->color(white)("dddddddddd")5y=-35)#
Divide both sides by 5
#color(green)(color(white)("dddddddddddddd")->color(white)("ddddddddddd")y=-7)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Check")#
#8x+5y=+21 color(white)("d")->color(white)("d") 8(7)+5(-7) =21#
#color(white)("dddddddddddd.d")->color(white)("dd")56-35 = 21color(red)(larr" True"#
#3x+5y=-14color(white)("d")->color(white)("d")3x+5y=-14#
#color(white)("dddddddddddddd")->color(white)("d")3(7)+5(-7)=-14color(red)(" True")#
