Posted by **Max** on Saturday, November 13, 2010 at 8:40pm.

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 -
**Damon**, Saturday, November 13, 2010 at 8:52pm
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 -
**Max**, Saturday, November 13, 2010 at 9:59pm
You subtract? arent you supposed to add?

- Algebra -
**Jen**, Saturday, November 13, 2010 at 10:06pm
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 -
**Max**, Saturday, November 13, 2010 at 10:15pm
THanks Damon and Jen! (especially Jen ;)

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