How do you solve #-x + y = 12# and #x + 2y = 3#?

2 Answers
Aug 1, 2015

Answer:

#{(x=-7), (y=5) :}#

Explanation:

All you really ahve to do to solve this system of equations is add the left side and the right side of the equations separately to eliminate the variable #x#.

#-color(red)(cancel(color(black)(x))) + y + color(red)(cancel(color(black)(x))) + 2y = 12 + 3#

#3y = 15#

To get the value of #y#, simply divide both sides of the equation by #2# to get

#(color(red)(cancel(color(black)(3))) * y)/color(red)(cancel(color(black)(3))) = 15/3 => y = color(green)(5)#

Now use this value of #y# in one of the two equations to get the value of #x#.

#x + 2 * 5 = 3#

#x + 10 = 3#

Add #-10# to both sides of the equation

#x + color(red)(cancel(color(black)(10))) - color(red)(cancel(color(black)(10))) = 3-10 => y = color(green)(-7)#

Aug 1, 2015

Answer:

I found:
#x=-7#
#y=5#

Explanation:

You can add the two equations (in column):
#{-x+y=12#
#{x+2y=3#
#color(red)(0+3y=15)#
so #y=15/3=5#
substitute this value into the first equatiom:
#-x+5=12#
#-x=7#
#x=-7#