# Algebra

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solving linear equations with 3 variables

1) x+y=1300
2) x+2z=1400
3) x+y+z=1600

this one should be so easy but i don't get which equation i should pick first?!

• Algebra -

It does not matter which you pick.
However since there is one times x in each it is easy to eliminate x.

x + y = 1300
x + 2z = 1400
-------------- subtract
y - 2 z = -100

then do the first and the third for example
x + y = 1300
x + y + z = 1600
---------------------
subtract
-z = -300
or z = 300
That was just lucky that there is no y term
If there were we would use the two equations left with y and z to get rid of one or the other.
now with z = 300 go back and use the one in x and z to get x
x + 2(300) = 1400
I think you can take it from there.

• Algebra -

You subtract? arent you supposed to add?

• Algebra -

A much easier way to do this would be:
1) x+y=1300
2) x+2z=1400
3) x+y+z=1600
Look at 1 and 2. Since x+y=1300 and x+y+z=1600, z obviously equals 300. Now substitute that back in #2.
x+2(300)=1400, x+600=1400, x=800.
Now substitute x and z back in #3.
x+y+z=1600, 800+y+300=1600, 1100+y=1600, y=500

• Algebra -

THanks Damon and Jen! (especially Jen ;)

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