Linear Equations- Solving for 3 Variables  TOPIC_SOLVED

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Linear Equations- Solving for 3 Variables

Postby Chiquita on Sat Aug 01, 2009 4:25 am

Hi everyone,
This is the problem I am having issues with:
4x + y - 3z = 11
2x - 3y + 27 = 9
x + y + z = (-3)
The instructions are to solve for x, y, and z.
I tried using two of the equations and eliminating one of the variables, then solving for one of the others. But then I got stuck- I didn't know if I should substitute that equation into one of the others... or do something else entirely! I'm not completely sure that I had started it off correctly, either. :(
I did read all of the lessons about Systems of Linear Equations and the Gaussian system but they only confused me more! Please help me...
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Postby stapel_eliz on Sat Aug 01, 2009 12:40 pm

There are any number of ways to proceed with the computations. One way might to be subtract the third equation from the first, eliminating "y". Then multiply the third equation by 3, and add to the second equation, again eliminating "y". This will give you two equations in just two variables, being "x" and "z".

Assuming the "27" in the second equation is a typo for "2z", you should be able to solve one of the new equations for "x=". Substitute the result into the other new equation, and solve for the value of "z". Back-solve for "x", and then again for "y". :wink:

If you get stuck, please reply showing your work and reasoning so far. Thank you! :D
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Re: Linear Equations- Solving for 3 Variables

Postby Chiquita on Sat Aug 01, 2009 7:13 pm

Thank you! I solved it correctly, and I got x= 2, y= (-3), and z= (-2). (Just in case you were interested in the answers) :wink:
Yes the "27" was a typo for "2z"... sorry about that.

I can't tell you how happy I am that I found this website- I can tell that it is going to be very helpful in Honors Pre-Cal next year!
Thanks again for helping me out! :D

~Chiquita~
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