Notice that the #y# terms both have the same coefficient.

Subtracting the equations will therefore eliminate the #y# terms.

#" "x" "color(red)(+" "5y)" " =-13" "............................ A#

#-6x" "color(red)(+" "5y)" " =8" ".......................... B#

Subtract #A - B" "# (change the signs of #B#)

#" "x" "+" "5y" " =-13" "color(white)(xxxxxxxxx)A#

#ul(color(blue)(+)6x" "color(blue)(-)" "5y" " =color(blue)(-)8)"
"color(white)(xxxxxxxxx)B#

#" "7xcolor(white)(xxxxxxxxx) =-21" "color(white)(xxxxxxxxxx)C"
"larr div 7#

#" "xcolor(white)(xxxxxxxxx) =-3#

Substitute #-3 " for " x# in #A#

#-3+5y =-13#

#" "5y = -13+3#

#" "5y=-10#

#" "y=-2#

An alternative method is to equate the #y# terms.

Transpose the equations to isolate #5y# in each:

#5y = -13-x" and "5y =8+6x#

We know that #" "5y=5y#

Therefore: #" "6x+8 = -13-x" "larr# now solve for #x#

#6x+x = -13-8#

#" "7x = -21#

#" "x =-3#

Then proceed as described above to get #y=-2#

Check in #B#

#-6x+5y#

#=-6(-3)+5(-2)#

#=18-10#

#=8" "larr# the answer is the same as the RHS