Algebra Know-It-ALL

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.

10. 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