Algebra Know-It-ALL

Part Two 339

Question 18-9

How can we be sure that the solution we have obtained for the three-by-three system pre-
sented in Question 18-1 is correct?

Answer 18-9

We must substitute our solutions into all three of the original equations. We did it indirectly
for the first equation in Answer 18-8, but if we want to be completely rigorous, we must plug
all three values into that equation along with the other two. (What if we made a mistake in
Answer 18-8? Don’t laugh. Things like that can and do happen!) Here are the original three
equations again, for reference:

4 x= 8 + 4 y+ 4 z

2 y= 5 +x− 5 z

4 z= 13 − 2 x+y

Grinding out the numbers in the first equation, we get

4 x= 8 + 4 y+ 4 z

4 × 6 = 8 + 4 × 3 + 4 × 1

24 = 8 + 12 + 4

24 = 24

Check one. In the second equation, we get

2 y= 5 +x− 5 z

2 × 3 = 5 + 6 − 5 × 1

6 = 5 + 6 − 5

6 = 6

Check two. In the third equation, we get

4 z= 13 − 2 x+y

4 × 1 = 13 − 2 × 6 + 3

4 = 13 − 12 + 3

4 = 4

Check three. Mission accomplished! Note that in the mixed addition/subtraction here, we
proceed directly from left to right after the equals sign. We subtract 12, and then we add 3.
We don’t subtract the quantity (12 + 3)! If we have any doubts when we come across a situa-
tion like this, we can change the subtraction to negative addition. If we do that in the above
calculation, we get an extra step, so the whole sequence goes like this: