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Coolmath Algebra
Systems of Equations 3x3's Lesson 2 - Solving by Elimination  
(page 3 of 8)

---- This algebra lesson explains how to solve a system of 3 equations and 3 unknowns by the elimination (addition) method.

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This one will be a little less finished...

5x - 2y + z = -3

              7y - 3z = -4

                4y + 3z = 26

Notice that we have a nice little 2 x 2 system that will give us the y and z answers...  Then, we'll just stick those into the first equation and we'll have x!

5x - 2y + z = -3
              7y - 3z = -4 and 4y + 3z = 26

Working the 2 x 2, eliminate the z...  to get y...  Then, use the y to pop
the
z...

( 7y - 3z = -4 ) + ( 4y + 3z = 26 ) = ( 11y = 22 ) ... y = 2 7y - 3z = -4 ... 7 ( 2 ) - 3 ( z ) = -4 ... 14 - 3z = -4 ... -3z = -18 ... z = 6

Now, stick y = 2 and z = 6 into the first equation to pop x:

5x - 2y + z = -3 ... 5x - 2 ( 2 ) + 6 = -3 ... 5x + 2 = -3 ... 5x = -5 ... x = -1

So, the answer is    ( -1 , 2 , 6 )  .

YOUR TURN:

       5x - 3y + 9z = -47

2x + 7y = 13

  4x + 5y = -1

* Notice that the 2 x 2 is x and y this time. 
You'll pop
z in the last step.


Continued on the next page

 The printing and distribution and/or downloading of these lessons is strictly prohibited.

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