Algebra II / Mr. Hansen
10/26/2005

Name: _________________________

Answers to Test on Chapters 3 and 4

 

1.(a)

S = {(–29/10, –15/4, –6/5, –71/20)}

 

 

(b)







The most common mistake that students made was in the notation used for writing the determinant. Remember, the determinant is a number, not a matrix. Matrices are surrounded by square brackets, but determinants use vertical bars as shown above.

 

 

(c)

24

 

Note: The easiest way to get this is to observe that y = Ny/D. Since we know what y and D are, namely –6/5 and –20, we can plug in to get the equation –6/5 = Ny/(–20). Solve this equation to get Ny = 24. A few students used this method to check their work.

The E period test used a different matrix with a different solution set, but strangely, the numerator determinant for y turned out to be 24 also.

 

 

2.(a)

–4 – (–16) = 12

 

 

(b)

12 – 12 = 0

 

 

3.

This is a matter of patience. Be sure to use the “question mark equal sign” when checking to see whether w + 2x + 4y – 4z really equals –1, and whether x – y – z really equals 1.

One way is to put w, x, y, and z all in terms of a common denominator of 20. Then we have w = –58/20, x = –75/20, y = –24/20, z = –71/20.

Equation 3: w + 2x + 4y – 4z = –58/20 + 2(–75/20) + 4(–24/20) – 4(–71/20)
       = (–58 – 150 – 96 + 284)/20
       = (–304 + 284)/20
       = –20/20
       = –1 ό

Equation 4: x – y – z = –75/20 – (–24/20) – (–71/20)
       = (–75 + 24 + 71)/20
       = (–75 + 95)/20
       = 20/20
       = 1 ό

 

 

4.

4

 

 

5.

Given line has slope –4/3. (A sketch shows this clearly. Please make a sketch!) Therefore, the ^ line has slope Ύ. Answer requires no additional work. Simply write the slope-intercept form:
       y + 1 = Ύ(x – 3)

Very few students got this one correct. You should probably go over this to make sure you understand the technique.

 

 

6.(a)

x = 3 [must show sketch for full credit]

 

 

(b)

y = –x – 2 [must show sketch for full credit]

 

 

(c)

y + p = 2(x – 1) [must show sketch for full credit]

 

 

(d)

x = 2 [must show sketch with vertical line passing through (2, 2.5)]

 

 

7.(a)

Most students got this, I am happy to say. There are vertical dotted lines passing through
(0, –2) and (0, 1.5), as well as a dotted line passing through (0, 2) and (2, 0). The upper region that is between the two vertical lines is the shaded solution set S. The remaining line, a solid line passing through (0, –4) and (4, 0), does not really play much of a role, since its only requirement is that the solution set must lie above, and S clearly does.

 

 

(b)

reduced row-echelon form

 

 

(c)

S = Ζ

 

 

(d)

S = {(x, y): x Ξ Β, y = 5 – x}
Or, you could say S = {(x, y): y Ξ Β, x = 5 – y} since that produces the same set of points.
Also acceptable: S = {the ordered pairs on the line y = 5 – x}.

 

 

(e)

S = {(5, –1)}