How do you solve the following system?: #-2x -5y =1, -x -y = -2#

1 Answer
Apr 13, 2016

Answer:

Just like this. It is a big answer but it is easy to understand

Explanation:

First, put one line above the other:
#-2x -y = 1#
#-x-y=-2#

Now, we check if one line may delete(remainder = 0) any element from the other adding them:
#-2x+(-x) = -3x=># No
#-y+(-y)=-2y=># No
#1 +(-2) = -1 =># No

Since add it has not deleted any element, multiply any of the lines by (-1) and try adding it again. I will chose the second line to multiply, its easier this way:
#-2x-y=1 ""=># #-2x-y=1#
#-x-y=-2 =># #x+y = 2#

Adding:
#-2x +x = -x =># No
#-y+y = 0 =># Yes!
#1 +2= 3=># No

Since this way one element was deleted (#y#), we may put it all in one "universal line":
#-x + 0 = 3# that becomes #x = -3#

To find #y# its way more simple, you just put the value of #x# in any place that he belongs to. Again I will chose the second line, still easier.
#-(-3) - y = -2#
#-y+3 = -2#
#-y = -5#
#y = 5#