You can use elimination or substitution.
I like to use the elimination method by making one of the variables into additive inverses. (In this case #color(red)(+2x and -2x)#)
This means that we can add 2 equations together to eliminate one of the variable.
#color(white)(xxxxxxxxxx)color(red)(2x)-5y =-4" "A#
#color(white)(xxxxxxxxxxx)x+6y =" "15" "B#
#B xx-2:" " color(red)(-2x)-12y = -30" "C#
#A+C:" "-17y = -34" "larr# now find #y#
#color(white)(....................................)y = 2#
Subst #2# for #y# in # B:" " x+6(2) = 15#
#color(white)(...................................)x+12 = 15#
#color(white)(....................................
........)x=3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You can also use substitution, by writing one variable in terms of the other.
#x +6y = 15" "rarr color(blue)(x = 15-6y)#
In #2color(blue)(x)-5y =-4" "larr# replace #x# by #color(blue)(15-6y)#
#2color(blue)((15-6y)) -5y= -4#
#30-12y -5y =-4#
#30+4 = 17y#
#34 = 17y#
#y =2#
Then #x = 15-6(2)#
#x=3#
