To solve a system using elimination, the co-efficients of one of the variable must be the same in both equations. It helps if they have opposite signs. Note that we have #+2x and -6x#

The #LCM# of #2 and 6# is #6#

#color(white)(xxxxxx)+2x+5y =5" "" "A#

#color(white)(xxxxxx)-6x+7y =-37" "B#

#A xx 3: rarr" "color(blue)(+6x)+15y =15" " " "C#

#color(white)(xxxxxxx.xx)color(blue)(-6x)+7y =-37" "B#

The #x# terms are now additive inverses. They add to #0#

#C+B: rarr color(white)(xxxxxx)22y =-22" "larr# no #x# term.

#color(white)(xxxxxxxxxxxxxxx) y = -1#

Substitute #-1# for #y# in any equation - let's use #A#

#color(white)(xxxxxx)2x+5(-1) =5" "" "A#

#color(white)(xxxxxxxx)2x" "-5 =5#

#color(white)(xxxxxxxxxxxxx)2x =10#

#color(white)(xxxxxxxxxxxxxx)x =5#

Check the values in another equation: Use #B#

#color(white)(xxxxxx)-6x+7y =-37" "B#

#color(white)(xxxxxx)-6(5)+7(-1) =-37#

#color(white)(xxxxxx)-30-7 =-37#

#color(white)(xxxxxxxxx)-37 =-37" "larr# the equation works out