How do you solve #x + y = -7# and #3x + y = -9#?

1 Answer
Aug 1, 2015

Answer:

#x = -1#; #y = -6#

Explanation:

You could solve this system of equations by using the multiplication method.

To do that, start by multiplying both sides of the first equation by #-3#.

#-3 * (x+y) = -3 * (-7)#

#-3x - 3y = 21#

The system of equations will now be

#{(-3x -3y = 21), (3x + y = -9) :}#

Next, add left sides and the right sides of the two equations separately to eliminate the terms that contain #x#

#-color(red)(cancel(color(black)(3x))) - 3y + color(red)(cancel(color(black)(3x))) + y = 21 - 9#

#-2y = 12#

Divide both sides of the equation by #-2# to get the value of #y#

#(-color(red)(cancel(color(black)(2)))y)/(-color(red)(cancel(color(black)(2)))) = 12/(-2) => y = color(green)(-6)#

Now that you know the value of #y#, use it in one of the two equations to determine the value of #x#

#3x + y = -9#

#3x + (-6) = -9#

This is equivalent to

#3x = -9 + 6 = -3#

Now divide both sides of this equation by #3# to get the value of #x#

#(color(red)(cancel(color(black)(3)))x)/color(red)(cancel(color(black)(3))) = (-3)/3 => x = color(green)(-1)#

The two solutions are

#{(x=-1), (y=-6) :}#