Fall-2012
LAGUARDIA COMMUNITY COLLEGE
Department of Mathematics, Engineering, and Computer Science
MATH096 ELEMENTARY ALGEBRA
Textbook: Beginning Algebra, sixth edition (EDUCO)
LAB # 4
Name: ____________________________________
Date: ____________________
Instructor: _________________________________
Section: __________________
You need to show all work. Indicate the right answer in the answer sheet. Even if you mark the right
answer, but do not show work on this sheet, you will not be given credit for that question:
1. (Tutor) Determine the equation of the line parallel to the line 2y = 3x – 9 and passing
through the point (– 6, 0).
a. 3y = 2x + 15
b. 2y − 3x = 9
c. 3x + 2y + 11 = 0
d. 3x = 2y − 18
e. 6x + 4y = 12
Solution: Writing in the form of y = mx + b.
�
�
�
y = �x − � → m = �
�
Slope of the parallel line= ;
�
passes through (−6,0)
Substitute these values in y = mx + b
�
0 = � ∙ (−6) + b → b = 9
The equation of desired line is:
�
y = � x + 9 → 3x − 2y = −18 (Ans.)
2. Find the equation of a line through the point (-1, 9) and parallel to 2x – 4y = 7.
3. (Tutor) Find the x value of the solution for the pair of equations:
2x – y = − 9
3x + 2y = 4
a. x = 0
b. x = −2
c. x = 2
d. x = 5
Revised: September, 2012
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M. Kothari
e. x = 3
Solution: Solving 2x − y = −9 for y → y = 2x + 9
Substitute 2x + 9 for y: 3x + 2y = 4 → 3x + 2(2x + 9) = 4
→ 3x + 4x + 18 = 4 → 7x + 18 = 4
→ 7x = −14 → x = −2
Therefore, x = −2
4. Find the y value of the solution for the pair of equations:
3x + 2y = 4
x+ y=1
5. (Tutor) Determine if the line is parallel, perpendicular, or neither.
4x + y = 13
y = -4x + 11
a. parallel
b. perpendicular
c. neiher
Solution: 4x + y = 13 → y = −4x + 13 → m� = −4
y = −4x + 11 → m� = −4
Since slopes of the two lines are equal, the lines are parallel.
6. Determine if the line is parallel, perpendicular, or neither.
-2x + 3y = -17
3x + 2y = -1
Revised: September, 2012
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M. Kothari
7. (Tutor) Determine the number of solutions the following system of equations has.
4x + y = 13
y = -4x + 11
a.No solution
b. Infinite solution
c. Unique solution
Solution: Rewrite both the equation in the form ax + by = c
4x + y = 13 → 4x + y = 13
y = −4x + 11 → 4x + y = 11
×(��)
4x + y = 13 �⎯⎯�
× (�)
4x + y = 11 �⎯�
−4x − y = −13
4x + y = 11
0x + 0y = −2 → 0 = −2 False
The statement is false. The system has no solution. It is inconsistent and consists of two
parallel lines.
8. Determine the number of solutions the following system of equations has.
2x -3y = 12
x + 2y = -1
9. (Tutor) Tracy invests $5,000, part at an annual rate of 5% and the remainder at an
annual rate of 7%. Her annual income from both investments is $310. How much did she
invest at each rate?
a)$1000 at 5% and 4000 at 7%
b)$2000 at 5% and $ 3000 at 7%
c)$1500 at 5% and $3500 at 7%
d) $ 2500 at 5% and 2500 at 7%
d) None of the above
Solution: Let the first investment amount = $𝑥 and the other amount = $𝑦
Step 1: The problem translates to:
x + y = 5000 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (I)
and
Revised: September, 2012
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M. Kothari
5
7
�
�x + �
� y = 310 → 5x + 7y = 31000 ⋯ (II)
100
100
Step 2: Write the first and the second equation together.
x + y = 5000
5x + 7y = 31000
Step 3: Eliminate x (or you can also chose to eliminate y) by multiplying the first
equation by -5 entirely. Then use the addition method to add two equations. This way
you will be eliminating the x.
−5x − 5y = −25000
5x + 7y = 31000
2y = 6000 → y = 3000
Substitute 3000 for y in equation x+y =5000→ 𝑥 = 2000
The first investment is of amount $2000 and second investment is of amount $3000
10. Jake borrowed $5000 from two different banks. One bank charges 7% simple interest,
and other charges 8.5% simple interest. If the total interest payment is $380, how much
did he borrow from each bank?
a)
b)
c)
d)
e)
$2500 at 7% and $2500 at 8.5%
$2000 at 7% and $3000 at 8.5%
$1000 at 7% and $4000 at 8.5%
$4000 at 7% and $1000 at 8.5%
$3000 at 7% and $2000 at 8.5%
11.(Tutor) How many pounds of Caramels costing $1.75 per pound must be mixed with
20 pounds of cream chocolates costing $2.50 per pound to produce a mixture costing
$2.25 per pound?
a) 6 lb. of caramel
b) 10 lb. of caramel
c) 8 lb. of caramel
d) 5 lb. of caramel
e) None of the above
Solution: Let the number of caramels=x pounds; number of chocolates= 20 pounds
The problem translates to:
1.75x + 20 × 2.50 = (x + 20) × 2.25 → 1.75x + 50 = 2.25x + 45
���
→ 1.75x + 50 = 2.25x + 45 �⎯⎯�= 175x + 5000 = 225x + 4500
→ 50x = 500 → x = 10 pounds
Revised: September, 2012
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M. Kothari
12. Caramels costing $ 3.75 per pound are mixed with chocolates costing $3
per pound to make a 6 pound mixture that will be sold at $ 3.25 per pound. How
many pounds of each are needed?
a)
b)
c)
d)
e)
3 lb. of caramel and 3 lb. of chocolates
2 lb. of caramel and 4 lb. of chocolates
4 lb. of caramel and 2 lb. of chocolates
1.5 lb. of caramel and 4.5 lb. of chocolates
2.5 lb. of caramel and 3.5 lb. of chocolates
-----------------------------------------------------------------------------------------------------------Extra Practice Problems (Optional)
1. Find the x value of the solution for the pair of equations:
2x – 3y = 18
3x + 2y = 1
2. Determine if the line is parallel, perpendicular, or neither.
3x + y = 5
-3x + y = 6
3. Determine the number of solutions the following system of equations has.
6x + 2y = 3
y = -3x + 3/2
4. X pounds of Caramels costing $1.75 per pound must be mixed with 20
pounds of cream chocolates costing $2.50 per pound to produce a mixture costing $
$2.25 per pound. Write an equation to determine the X.
5. There are 40 students in Professor Kay’s math class. Some students will take Test A
and the others will take Test B. The number of students taking Test A is five less than
twice the number taking Test B. How many students will take each test?
a) Test A = 20 and Test B =20
b) Test A = 15 and Test B = 25
c) Test A = 25 and Test B = 15
d) Test A = 22 and Test B =18
e) Test A = 35 and Test B = 5
Revised: September, 2012
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