Part Two 329Answer 16-6
The solution we have obtained is x=−3 and y= 2. We must plug these values into both of the
original equations to be certain we’ve gotten the right solution. For the first original equation,
we proceed like this:
2 x−y+ 8 = 0
2 × (−3)− 2 + 8 = 0
− 6 − 2 + 8 = 0
0 = 0
It checks out! For the second original equation, we do this:
x− 3 y+ 9 = 0
− 3 − (3 × 2) + 9 = 0
− 3 − 6 + 9 = 0
0 = 0
It checks again! Now we know the solution we obtained is correct.
Question 16-7
Do you suspect that I concocted the above problem so it would come out with a pair of “clean
integers” for the solution? If so, you are right! How can we compose a two-by-two linear sys-
tem as a test problem (for someone else to solve), and be sure the solution will turn out to be
a pair of integers?
Answer 16-7
We can choose a point where the graphs of two lines intersect, and assign different slopes to
those lines. Then we can write down the equations in point-slope form, using the solution
point as the reference for both lines. Finally, we can convert the point-slope equations to some
other form to get the test problem. For extra credit, you can try this and then solve the test
problem you’ve created.
Question 16-8
How can we add multiples of the two original equations stated in Question 16-2 to solve the
linear system for x? For reference, here are the equations again:
2 x−y+ 8 = 0
x− 3 y+ 9 = 0Answer 16-8
We can multiply the first equation through by −3 and then add it to the second equation,
getting the sum
− 6 x+ 3 y− 24 = 0
x− 3 y+ 9 = 0
− 5 x− 15 = 0