Algebra Know-It-ALL

Part Two 337

We can solve this system in many different ways. In Chap. 18, we learned to solve systems of
this type by getting rid of one variable, resulting in a two-by-two linear system, solving that
system, and then substituting back to solve for the variable we eliminated. That’s the method
we’ll use here. We can get rid of any of the three variables by morphing and adding any two
pairs of the three-variable equations.

Question 18-5

How can we obtain a two-by-two linear system in x and z from the three-by-three system as
stated in Answer 18-4, using the first two equations and then the second two?

Answer 18-5

Here are the first two equations from the three-by-three linear system as stated in Answer 18-4:

4 x− 4 y− 4 z= 8

−x+ 2 y+ 5 z= 5

We can divide the top equation through by 2 and then add the bottom equation, getting
the sum

2 x− 2 y− 2 z= 4

−x+ 2 y+ 5 z= 5

x+ 3 z= 9

That’s the first equation in our two-by-two system. To get the second equation, let’s look at
the second two equations from the three-by-three system as stated in Answer 18-4:

−x+ 2 y+ 5 z= 5

2 x−y+ 4 z= 13

We can multiply the bottom equation through by 2 and then add it to the top equation,
getting

−x+ 2 y+ 5 z= 5

4 x− 2 y+ 8 z= 26

3 x+ 13 z= 31

Now we have the following two-by-two linear system in the variables x and z :

x+ 3 z= 9

3 x+ 13 z= 31

Question 18-6

How can we solve the above two-by-two system for z using the addition method?