Algebra Know-It-ALL

Let’s multiply the second equation through by −3, getting

− 6 x+ 15 y+ 3 z= 3

If we add this to the first equation, we obtain the sum

− 4 x+ 2 y− 3 z= 5

− 6 x+ 15 y+ 3 z= 3

− 10 x+ 17 y= 8

Make z vanish again

Now let’s scrutinize the second two revised equations. Here they are again:

2 x− 5 y−z=− 1

and

3 x+ 6 y− 7 z= 0

We can multiply the first equation through by −7, getting

− 14 x+ 35 y+ 7 z= 7

When we add the second equation to this, we get

− 14 x+ 35 y+ 7 z= 7

3 x+ 6 y− 7 z= 0

− 11 x+ 41 y= 7

State the two-by-two

We have now derived two different equations in the variables x and y. This is a two-by-two
linear system, which we know how to solve:

− 10 x+ 17 y= 8

and

− 11 x+ 41 y= 7

Are you confused?

When we set out to get rid of the variable z, we decided to work on the first two equations in the revised
three-by-three system, and then work on the second two equations. You might ask, “Why can’t we elimi-
natez between the first equation and the second one, and then between the first equation and the third

Eliminate One Variable 283