Algebra Know-It-ALL

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

and

y= (−2/3)x− 2

There’s no morphing for us to do. So let’s go ahead and mix the right sides of these two

equations. We get

(4/3)x+ 4 = (−2/3)x− 2

We can multiply this equation through by 3 to obtain

4 x+ 12 =− 2 x− 6

Adding 2x to each side gives us

6 x+ 12 =− 6

Subtracting 12 from each side, we get

6 x=− 18

Dividing through by 6 tells us that x=−18/6=−3. We can plug this value for x into

either of the SI equations to solve for y. Let’s use the first equation. We have

y= (4/3)x+ 4

= (4/3) × (−3)+ 4

=−12/3+ 4

=− 4 + 4

= 0

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

This represents the point where lines L and M intersect in Fig. 17-9.

4. Figure B-8 shows the transformation process, one step at a time, exactly as I did it
   using the rotate and mirror functions in my computer graphics program. At A, we see
   the original graph, identical to Fig. 17-9. At B, the entire coordinate grid, the lines,
   and the points have been rotated counterclockwise, all together, by 90°. Even the label
   characters have been rotated! At C, the graph from B has been mirrored. Even the label
   characters have been reversed! (That looks strange, but it can help us see what is taking
   place.) At D, everything has been relabeled to conform to the new coordinate system.
   The transposed lines are called L and M. The numbers in the ordered pairs have been
   transposed, because y is now the independent variable and x is the dependent variable.


5. In part D of Fig. B-8, line L passes through (0, −3) and (4, 0). The x-intercept
   is −3. Remember that x is now the dependent variable, so it’s the x intercept, not
   they-intercept, that concerns us. We can travel along L from (0, −3) to (4, 0)