Equation 1:
Substitute #color(red)(2)# for #color(red)(x)# and substitute #color(blue)(-2)# for #color(blue)(y)# in the first equation and determine if the equation is true:
#3color(red)(x) + color(blue)(y) = 4# becomes:
#(3 xx color(red)(2)) + color(blue)((-2)) = 4#
#6 - 2 = 4#
#4 = 4#
Because this equation is true we can move to the next equation.
Equation 2:
Again, Substitute #color(red)(2)# for #color(red)(x)# and substitute #color(blue)(-2)# for #color(blue)(y)# this time in the second equation and determine if the equation is true:
#color(red)(x) + 3color(blue)(y) = -4# becomes:
#color(red)(2) + (3 xx color(blue)(-2)) = -4#
#color(red)(2) + (-6) = -4#
#color(red)(2) - 6 = -4#
#-4 = -4#
Because the first and second equation are both true for the point in the problem the point or ordered pair IS a solution for the system of equations.