Solve the equations #-x-y=11# and #-x+3y=-13# by elimination method?

1 Answer
May 25, 2017

#x=-5# and #y=-6#

Explanation:

In elimination method what we try to do is eliminate one of the variable, whether #x# or #y#.

Using #x-# Observe that coefficient of #x# in both the equations is same i.e. #-1#. Hence if we subtract one equation from other #x# will get eliminated.

Therefore #-x-y-(-x+3y)=11-(-13)#

or #-x-y+x-3y=11+13#

or #-cancelx-y+cancelx-3y=24#

or #-4y=-24# i.e. #y=24/(-4)=-6#

Putting this in first equatin we get #-x-(-6)=11#

or #-x+6=11# i.e. #-x=11-6=5# or #x=-5#

and solution is #x=-5# and #y=-6#

Using #y-# Observe that coefficient of #y# in both the equations is different. While in first equation it is #-1# in second equation it is #3#. Hence we multiply first by #3# and add it to second equation and #y# will get eliminated.

#3xx(-x-y)+(-x+3y)=3xx11+(-13)#

or #-3x-3y-x+3y==20#

or #-3x-cancel(3y)-x+cancel(3y)=20#

or #-4x=-20# i.e. #x=20/(-4)=-5#

Putting this in first we get #-(-5)-y=11#

i.e. #5-y=11#

or #-y=11-5=6# i.e. #y=-6#