Solve by elimination and substitution
#1)color(blue)(-2x+5y=-3#
#2)color(blue)(x-3y=-2#
If you see carefully, you could eliminate #-2x# from the first equation by #x# in the second equation if we multiply #x# with #2# to get #2x#
#rarr2(x-3y=-2)#
Use distributive property #color(brown)(a(b+c=d),ab+ac=ad#
#rarr2x-6y=-4#
Now, add both of the equations
#rarr(-2x+5y=-3)+(2x-6y=-4)#
#rarr-y=-7#
#rArrcolor(green)(y=7#
Substitute the value of #y# to the first equation
#rarr-2x+5(7)=-3#
#rarr-2x+35=-3#
#rarr-2x=-3-35#
#rarr-2x=-38#
#rArrcolor(green)(x=(-38)/-2=38/2=19#
Check (Substitute the values of #x# and #y# to the first equation)
#color(orange)(-2(19)+5(7)=-3#
#color(orange)(-38+35=-3#
#color(orange)(-3=-3# =) correct!
Now to the second equation
#color(indigo)(19-3(7)=-2#
#color(indigo)(19-21=-2#
#color(indigo)(-2=-2# :) correct!
#:.# #x and y# have values of #19 and 7#
