Algebra Know-It-ALL

Part Two 335

Answer 17-7

They are the equations we derived in Answers 17-3 and 17-5:

x= (1/2)y− 4

and

x= 3 y− 9

Question 17-8

What is the x-intercept of the line representing the equation x= (1/2)y− 4?

Answer 17-8

We can answer this straightaway, because the equation is in SI form with x as the dependent
variable. It’s −4.

Question 17-9

What is the x-intercept of the line representing the equation x= 3 y− 9?

Answer 17-9

Again, we can infer this from the SI equation having x as the dependent variable. It’s −9.

Question 17-10

Based the known slopes of the lines, and on the point data shown in Fig. 20-7, at what points
would the lines representing the two-by-two linear system intersect the graph of the equation
y=−2?

Answer 17-10

The graph of the equation y=−2 would appear as a vertical line in Fig. 20-7, parallel to the
x axis and running through the point (−2, 0). To reach this line from the point (2, −3), we can
travel in the negative y direction by 4 units along either of our existing lines. First, let’s move
along the line for x= (1/2)y− 4. The slope is 1/2. That means if we go 4 units in the negative
y direction, we must go 4/2, or 2, units in the negative x direction to stay on the line. That will
put us at the point (−2,−5). Now let’s move along the line for the equation x= 3 y− 9. The
slope is 3. Therefore, if we go 4 units in the negative y direction, we must go 4 × 3, or 12, units
in the negative x direction to stay on the line. That will get us to the point (−2,−15).

Chapter 18

Question 18-1

Here is a set of equations that forms a three-by-three linear system:

4 x= 8 + 4 y+ 4 z

2 y= 5 +x− 5 z

4 z= 13 − 2 x+y