How do you solve the system #x + y = 2# and #2x + y = -1#?

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Answer 1 · 2018-04-05T10:02:37

#x=1/3#

#y=5/3#

from this equation

#x+y=2#

#x=2-y#

put the value of #x# in this equation

#2x+y=-1#

#2(2-y)+y=-1#

#4-4y+y=-1#

#-3y=-5#

#y=5/3#

#x+y=2#

#x+5/3=2#

#x=1/3#

Answer 2 · 2018-04-05T10:05:59

#x = -2, y = 3#

This is a simultaneous equation question. In the interest of time, I will solve it using the elimination method:

#x + y = 1#

#2x + y = -1#

To solve using elimination, all we do is literally subtract the bottom line from the top line which gets rid of the y #(y-y = 0)#, giving us:

#-x = 2#

#x = -2#

We then substitute (#x =-2#) into either of the equations:

#x + y = 1#

#(-2) + y = 1#

#y = 3#

And then check that this works, using substituting both values into the second equation.

#-2x + y = -1#

#2(-2) + (3) = -1#

#-1 =-1#, so we know the values are correct!