# Linear Equations through Elimination?

• Aug 10th 2009, 07:45 AM
Linear Equations through Elimination?
NO idea how to do this. Can someone please help me?

{-x + 3y = -1
x- 2y = 2

(Wink)
• Aug 10th 2009, 07:54 AM
y=1

x=4
• Aug 10th 2009, 08:06 AM
???
How did you get that?
• Aug 10th 2009, 08:10 AM
skeeter
Quote:

Originally Posted by SoobadatMath.
NO idea how to do this. Can someone please help me?

{-x + 3y = -1
x- 2y = 2

(Wink)

add the two equations, term for term ...

-x + 3y = -1
x - 2y = 2
------------
0 + y = 1 ... note that the x terms cancel

now sub 1 in for y in either equation and solve for x.
• Aug 10th 2009, 08:21 AM
So then x + 3y = -4
y + x = 0
???????????????????
• Aug 10th 2009, 08:44 AM
Wilmer
Quote:

Originally Posted by SoobadatMath.
So then x + 3y = -4
y + x = 0
???????????????????

What are you doing ? (Shake)

You were told that y = 1.
So replace the y with 1 in one of the equations, then solve for x.
• Aug 10th 2009, 08:58 AM
skeeter
Quote:

Originally Posted by SoobadatMath.
So then x + 3y = -4
y + x = 0
???????????????????

new system ...

x + 3y = -4
y + x = 0

multiply every term in the second equation by -1 ...

x + 3y = -4
-x - y = 0
-------------

add 'em up and solve.
• Aug 10th 2009, 09:08 AM
Wilmer
x + 3y = -4 [1]
y + x = 0 [2]
If that's a new system, then from [2]: y = -x
So [1]: x - 3x = -4