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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