Download Math 60 Test # 2 Fall 2014

Math 60 Test # 2 Fall 2014
Chapter 3-4
Instructor: Smith-Subbarao
_______Solutions__________________
Name
Score: _______________
Percent: ______________
Directions:
• Show all work and circle/box your answers.
•
Partial credit may be given, even if the answer is incorrect, if your work is clear – attach
additional scratch pages you wish to be considered.
•
If you do not show your work, you may not get credit.
•
Unless otherwise instructed, leave all answers as fraction; improper fractions are OK.
•
Not all problems are worth the same number of points
NO: Telephones, Books, Notes; CALCULATORS ARE NOT ALLOWEED
Suggestions:
• Choose the problems you understand best to work first.
• If you get stuck, write down what you do understand for partial credit and move on
• Show your work clearly
• Check your solutions
• Evaluate your solutions for “reasonableness”
1. Find the slope, y intercept, and equation of the line that goes through the points
(-3, 1) and (-1, 3) and graph it on the coordinate plane. (15 points)
Slope
__1_______
Intercept ____(0, 4)_______
Equation _____y = x + 4_______
11
10
9
8
7
6
5
4
Line 1
Line 2
y
3
2
1
0
-7
-6
-5
-4
-3
-2
-1 -1 0
1
2
3
4
5
6
7
-2
-3
-4
-5
x
Then, make a table of values and graph the following equation:
y = -x + 2
Find a point that solves both equations.
(-1, 3)
Scoring: slope 2pts, intercept 2 pts, equation 2 pt, graph part 1 2 pts, table 2 pts, 2nd graph 2
pts, solution 3 pts. Note, the intersection of the two lines is the solution – you should have put
them on the same graph.
2. Find the solution to the following system of equations by substitution: (10 pts)
3x – z = 4x
2z + 4 = x + 1
The first becomes –z = x, substituting the second is
-2x + 4 = x + 1
-3x = -3
x=1
z=1
To get full credit you had to use the method of substitution.
3. Determine whether the given pairs of lines is parallel, perpendicular, or neither. (10 pts)
a. x – 4y = 5
b.
8x – 5y = 10
y – 8 = 1 - 4x
3x = 35 – 3y
y = x/4 – 5/4
y = 8/5 x – 2
y = -4x + 9
y = -x – 35/3
Slopes are negative reciprocals
Slopes have no relationship
Perpendicular
Neither
4. Is the ordered pair (1, -2) a solution to the following systems of equations? (10 pts)
a. 3x – y = 5x
y + 8 = 2(x+ 2)
b.
10x – 5 = -5(y+ 1)
x=y–1
3(1) – (-2) = 5(1)
10(1) – 5 = -5(-2 + 1)
3 + 2 = 5 OK
10 – 5 = 5 OK
(-2) + 8 = 2 (1 + 2)
1 = (-2) – 1
6 = 6 OK
1 = -3 Not a solution
Is a solution for both, so is solution
Is not a solution of the pair
5. Find the solution to the following system of equations by addition (elimination): (10 pts)
5x – 3y = 10
3x – 2y = 5
Multiply the first equation by 2 and the second by 3
10x – 6y = 20
9x – 6y = 15
Subtract the 2nd from the 1st
x=5
substitute to find y into the 1st:
25 – 3y = 10, y = 5
Y = 5, y = 5
Find the solution to the systems of equations in problems 6 and 7 using any method,
(5 pts each)
6. x – 2y = 10
4y = 2(x – 2) + 2 = 2x – 4 + 2 = 2x - 2
From the first, x = 10+2y, substituted in the second
4y = 2(10 + 2y) – 2
4y = 20 + 4y – 2
0 = 18 NO SOLUTION is possible
7. x/2 + y/3 + 1/2 = 2
x + 2y = 3
Multiply the first by 6
3x + 2y + 3 = 12 or 3x + 2y = 9
x = 3 – 2y from the second, substituted into the equation above
3(3 – 2y) + 2y = 9
9 – 6y + 2y = 9
y=0
x = 3 -2y = 3
x = 3, y = 0
8. Two angles are complimentary. One is four times larger than the other. What are the
measures of the two angles? (5 pts)
x = 4y
x + y = 90 since they are complimentary angles
4y + y = 90
y = 18, x = 72
9. Solve for x: (5 pts)
5(x+1) – 3x < 4x – 1
5x + 5 – 3x < 4x – 1
2x + 5 < 4x – 1
6 < 2x
x>3
10. A student solved |3x – 5| = 22 as follows: (5 pts)
|3x – 5| = 22
3x - 5 = 22
3x = 27
x=9
What errors, if any, did he make? If none, write none. If he erred, give the correct solution
Need to also solve 3x – 5 = -22
3x = -17, x = -17/3
x = 9 or x = -17/3
to get full credit you needed to find the second solution
11. Solve for x: (10 pts)
x/3 < x/2 + 1< 2x
Break into two inequalities:
x/3 < x/2 + 1 and x/2 + 1 < 2x
For the first, multiply by 6
2x < 3x + 6, or x > -6; if you got to –x < 6, you needed to change the orientation if you
divided by a negative number.
For the second, multiply by 2:
x + 2 < 4x
3x > 2, x > 2/3
We have x > -6 and x > 2/3. The only values which satisfy all inequalities are x > 2/3. If
you did not come to that conclusion, you lost a point
12. Karen invested part of her $10,000 in savings bonds at 5% simple interest and the rest in
stocks at 7% simple interest. If she receives $650 a year in interest, how much did she
invest in each account? (10 pts)
Let S be the amount is stocks, B the amount in bonds
S + B = 10,000
0.05 S + 0.07 B = 650
Multiply by 100:
5S + 7B = 65000
From S + B = 10,000, we have S = 10,000 – B; substitute
5(10,000 – B) + 7B = 65000
50,000 + 2B = 65000
2B = 15000, B = 7500
10,000 – 7500 = 2500 = S