The equations can be solved by the elimination method or the substitution method. The elimination method has been used by Jade.
Let's use the substitution method because one of the equations has a single variable.
#color(blue)(-x) +4y = 24" "# can be re-arranged to give: #color(blue)((4y-24) = x#
This means that #color(blue)(x and (4y-24))# are interchangeable.
We have 2 equations with 2 unknowns:
#" "6y +3color(blue)(x) = 18" and "color(blue)(x = 4y-24)#
#color(white)(xxxxxxx)darr#
#6y +3color(blue)((4y-24)) = 18" "larr# replace #x# with #(4y-24)#
#6y + 12y -72 = 18" "larr# an equation with only y, solve it
#18y =18+72#
#18y = 90#
#color(red)(y = 5)" but "x = 4color(red)(y)-24#
#color(white)(xxxxxxxxxxxxxxxxxxxx)x = 4color(red)((5))-24#
#color(white)(xxxxxxxxxxxxxxxxxxxx)x = 20-24#
#color(white)(xxxxxxxxxxxxxxxxxxxx)x = -4#
#x = -4 and y = 5#
