644 Worked-Out Solutions to Exercises: Chapters 11 to 19
and simplified to
x+x+ 83 = 13
When we subtract 83 from each side and note that x+x= 2 x, we get
2 x=− 70
Therefore, x=−70/2=−35. We can replace x with −35 in the SI equation above to solve
fory, getting
y=−x− 83
=−(−35)− 83
= 35 − 83
=− 48
When we check back and compare this with solution to Prob.4, we see that the answers
are the same: x=−35 and y=−48.
- Here are the equations again, for reference:
s= 2 r− 3
and
− 10 r+ 5 s+ 15 = 0
This pair of equations appears well suited to a solution by substitution. The first equation
gives us s directly in terms of r. Let’s replace s by (2r− 3) in the second equation. We get
− 10 r+ 5(2r− 3) + 15 = 0
The distributive law allows us to morph the left side of this equation into a straight sum:
− 10 r+ 10 r− 15 + 15 = 0
which simplifies to
0 = 0
This statement is true, so we can’t claim a contradiction. But it’s useless for solving this system.
(If we try any other method to solve it, we’ll encounter a similar barrier.) The trouble becomes
clear if we solve the second original equation for s directly in terms of r. We start with
− 10 r+ 5 s+ 15 = 0