Algebra Know-It-ALL

648 Worked-Out Solutions to Exercises: Chapters 11 to 19

Now that we know the slope and the x-intercept for line M*, we can write its SI equation as

x= (−3/2)y− 3

7. We have the equations for lines L and M from part D of Fig. B-8 in SI form.
   Together, they constitute a two-by-two linear system:

x= (3/4)y− 3

and

x= (−3/2)y− 3

There’s no morphing to do here; these equations are ready to mix. When we combine the

right sides, we get

(3/4)y− 3 = (−3/2)y− 3

We can multiply this through by 4 to obtain

3 y− 12 =− 6 y− 12

Adding 12 to each side gives us

3 y=− 6 y

When we add 6y to each side, we get

9 y= 0

Dividing through by 9 tells us that y= 0. We can plug this value for y into either of the SI

equations to solve for x. Let’s use the first equation. We have

x= (3/4)y− 3

= (3/4) × 0 − 3

= 0 − 3

=− 3

Having found the solution y= 0 and x=−3, we can state it as the ordered pair (0,−3).

This represents the point where lines L* and M* intersect in part D of Fig. B-8. Remem-

ber that this ordered pair is of the form (y,x), not (x,y). That’s because in this situation,

y is the independent variable and x is the dependent variable.

8. If a linear function has a graph with a slope of 0, then the inverse relation is not
   a function. That’s because the graph of the inverse relation is a line parallel to the
   dependent-variable axis. The domain of that inverse relation is a single number, and