Mat 117 Exam 1
1. [6 pts each] Find the real solutions of the following equations: Show your work.
Give exact answers.
1
3
(a) ( x  2)  ( x  3)  5
4
2
Solution:
The first thing to do is deal with the fraction. I will do it by multiplying both sides by 4.
3
1

4 ( x  2)  ( x  3)   4(5)
2
4

Distribute the 4 through on the left-hand side.
1
3
(4) ( x  2)  (4) ( x  3)  20
4
2
( x  2)  6( x  3)  20
Now distribute the 1, combine like terms, and isolate x.
x  2  6 x  18  20
 5 x  20  20
 5 x  40
40
x
 8
5
2
x

40
x2 x2
Solution:
The first thing to do is deal with the fraction. I will do it by multiplying both sides
by x  2 .
2
 x

( x  2)

 4  0
x2 x2

Distribute the x  2 through on the left-hand side.
x
2
( x  2)
 ( x  2)
 ( x  2)4  0
x2
x2
x( x  2) 2( x  2)

 (4 x  8)  0
x2
x2
The x  2 factors in the numerator and denominator of the first two terms cancel. This
leaves us with
x  2  (4 x  8)  0
Now distribute the 1, combine like terms, and isolate x.
x  2  4x  8  0
(b)
 3 x  10  0
 3 x  10
10
10
x

3
3
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Mat 117 Exam 1
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2. [6 pts] The price of a new car including a 5.5% sales tax is $16,568.80. What was the
price of the car before the sales tax was added? Show your work. (Round your
answer to two decimal places)
Solution:
The thing we don’t know is the price of the car, P. In order to calculate the amount paid,
we must compute the sales tax by multiplying the tax rate (in decimal form) by the price
of the car ( .055P ). We now add the sales tax to the price of the car to get the total paid.
P  .055P  16568.80
We now solve the equation for P to get the original price of the car before tax.
1.055P  16568.80
16568.80
P
1.055
P  15705.02369668
Thus the original price of the car was $15,705.02.
3. [6 pts] Solve the quadratic equation by factoring: 12 x  5  4 x 2 Show your work.
Solution:
This is a quadratic equation. We first need to put it in standard form.
4 x 2  12 x  5  0
We now need to determine the factors. Since the constant term is 5, there are only two
possible factors (ignoring signs) 5 and 1. Keeping in mind that the leading coefficient is
4, there does not seem any clear way to put together the 4 and 1 factors of 4 with the 5
and 1 factors of 3 to produce a 12. Thus we want to try the other set of factors for 4
with are 2 and 2. Using these we get
(2 x  5)( 2 x  1)  0
Now set each factor equal to zero and solve.
2x  5  0
2x  1  0
2x  5
x
5
2
2x  1
1
x
2
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Mat 117 Exam 1
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4. [6 pts] Use your calculator to solve:
2.6 x 2  5.6 x  1.5  0
Round your answers to three decimal places. Show your work and explain how
you used your calculator.
Solution:
Since I am told to use a calculator, I can solve this by either graphing the equation or by
using the quadratic formula.
Graphing
Quadratic Formula
To solve this graphically, the first thing I
To use the quadratic formula, we need to
need to do is graph the equation. The
determine a, b, and c for the equation.
graph in the standard window looks like
a = 2.6, b = 5.6, and c = 1.5
Now we plug into the quadratic formula
right
and do the basic computations on the
solution
calculator.
left solution
The points where this graph hits the x-axis
are the solutions to the problem. To find
these point, you can use the calculate
feature on the calculator. To do that you
press 2nd and Calc. Then you choose the
option zero (root). You are then asked for
a left bound, a right bound and a guess.
For the left-most solution you get x =
.240912. You repeat the process for the
solution on the right and you get x =
2.39475. We need to round to 3 decimal
place, so the answers are x = 2.395 and x
=.241.
x
 b  b 2  4ac
2a
x
5.6  (5.6) 2  4(2.6)( 1.5)
2(2.6)
5.6  31.36  15.6
5.2
5.6  46.96 5.6  6.85274
x

5 .2
5.2
5.6  6.85274 12.45274
x

 2.39476
5 .2
5.2
5.6  6.85274  1.25274
x

 .240912
5.2
5.2
We need to round to 3 decimal place, so the
answers are x = 2.395 and x =.241.
x
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Mat 117 Exam 1
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5. [6 pts each] Solve the following equations algebraically. Show your work. Give
exact answers.
(a) x 3  x 2  4 x  4  0
Solution:
To solve this equation, we first need to look at it as two parts x 3  x 2 and  4x  4 and
factor each part where possible. The first part factors as x 2 ( x  1) . The second part
factors as  4( x  1) . We chose 4 since we wanted to end up with ( x  1) as one of the
factors as we did with the first part. Now we put the two factored parts back together into
an equation and we have x 2 ( x  1)  4( x  1)  0 . Now this equation can be factored
since each term has a factor of x  1 . When we factor, we get ( x  1)( x 2  4)  0 . We
can factor further since the second factor x 2  4 is the difference of two squares. We
then end up with ( x  1)( x  2)( x  2)  0 . This is completely factored and we can now
use the zero product principle, which allows us to set each factor to zero, to finish solving
the equation. So we have
x 1  0
x20
x20
x 1
x2
x  2
(b) 4 x  3  9
Solution:
To solve this equation, we need to get rid of the square root. In order to do that, we will
square both sides of the equation and then solve as usual.
4x  3  92
4 x  3  81
4 x  78
78
x
 19.5
4
(c) | 5 x  2 | 10
Solution:
When solving an equation with an absolute value sign, we need to take into account that
the expression within the absolute value sign could be either positive or negative. In this
case the expression 5x  2 could either be 10 or 10. This means we need to solve two
different equations.
5 x  2  10
5 x  2  10
5x  8
5 x  12
8
 12
x
x
5
5
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Mat 117 Exam 1
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(d) ( x  5) 5 / 3  32
Solution:
This problem is much like part (b). We need to raise both sides of the equation to a
power so that the exponent on the left-hand side will be one. To do this we use the power
3
. Keep in mind that when you raise an expression raised to an exponent to another
5
exponent, the exponents are multiplied. Thus we have
3
5  5
3

 ( x  5) 3   (32) 5




 *
( x  5) 3
5
3
5
8
( x  5)1  8
x5 8
x3
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Mat 117 Exam 1
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6. [6 pts] Find the domain of the expression x 2  x  30 either graphically or
algebraically. Show your work and/or explain how you used your calculator.
Solution:
To solve this problem we need to solve the inequality x 2  x  30  0 since the domain
of a square root is any number which is positive or zero.
Graphically
Algebraically
To solve the inequality graphically, we
Since this is a quadratic inequality, I will first
need to graph the left-hand side of the
need to solve the quadratic equation
inequality on our graphing calculator. We
x 2  x  30  0 to determine the critical points.
do not need to graph the right-hand side
We can see that it factors into
since it is represented by the x-axis. In the
( x  5)( x  6)  0 . If we set each factor equal to
standard window, we see:
zero and solve, we get the critical points
x5  0
x60
and
. Now to solve the
x5
x  6
inequality, we need to test a point which is less
than 6 (say 7), a point which is between 6
and 5 (say 0) and a point that is greater than 5
(say 6). When we plug 7 into the original
Although this is not a complete picture of
(7) 2  (7)  30  0
the graph, we can still use it for our
inequality, we get 49  7  30  0
which
purposes here. It is now possible to tell
12  0
that the parabola crosses the line y  0 (xis true, thus numbers less than or equal to 6 are
axis) in two places. The problem asks,
where the parabola is above or intersecting part of the answer. When we plug 0 into the
the line (x-axis). To calculate these two
points where the parabola and the line
intersect, we use 2nd, Calc, zero (root). The
left-most point is (6,0) and the right-most
point is (5,0) . The parabola is above the
line (x-axis) for x values to the left of 6
and for x values to the right of 5. Thus the
solution to the inequality is
(,6]  [5, ) .
2
original inequality, we get 0  0  30  0
 30  0
which is not true, thus number between 6 and
5 are not part of the answer. When we plug in 6
to the original inequality, we get
6 2  6  30  0
36  6  30  0 which is true, thus numbers
12  0
greater than or equal to 5 are part of the answer.
In summary the solution to the inequality is
 ,6  5,  .
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Mat 117 Exam 1
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7. [6 pts each] Solve the following inequalities algebraically (Show your work and
sketch the graph of the solution set):
(a) 3x  1  7
Solution:
To solve this absolute value inequality, we need to realize that in order for the inequality
to be true the expression 3x  1 can only have values between and including 7 and 7.
To express this in a mathematical expression we have  7  3x  1  7 . We now solve
this 3-part inequality.
 7  3x  1  7
 8  3x  6
8
x2
3
We now need to sketch a graph of the solution set.
8
8
2
2
3
3
3
0
3
or
3
[
]
0
3
4x
 4 x
5
Solution:
To solve this inequality, I will first multiply both sides by 4. Since 4 is positive, this will
not switch the sign of the inequality.
4x 

5 2    5(4  x)
5 

20 x
10 
 20  5 x
5
10  4 x  20  5 x
(b) 2 
10  20  5 x  4 x
 10
Grouping like terms, I get:  10  9 x
or x 
.
9
 10
x
9
We now need to sketch a graph of the solution set.
 10
 10
9
9
3
0
3
or
3
)
0
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Mat 117 Exam 1
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x3
0
x2
Solution:
To solve this inequality we need to find the critical points, which are the x-values that
make either the numerator or the denominator equal to zero. The critical point from the
x3 0
x20
numerator is
and the critical point from the denominator is
. Now to
x  3
x  2
solve the inequality, we need to test a point which is less than 3 (say 4), a point which
is between 3 and 2 (say 2.5) and a point that is greater than 2 (say 0). When we
43
0
42
1
plug 4 into the original inequality, we get
which is true, thus numbers less
0
2
1
0
2
than or equal to 3 are part of the answer. The number 3 is included since it makes the
fraction equal to zero and thus satisfies the inequality. When we plug in 2.5 to the
 2 .5  3
0
 2.5  2
.5
0
original inequality, we get
which is not true, thus number between 3 and
 .5
1  0
2 are not part of the answer. When we plug 0 into the original inequality, we get
03
0
02
which is true, thus numbers greater than 2 are part of the answer. The
3
0
2
number 2 is not part of the answer since it would cause the denominator to be zero and
division by zero is undefined. In summary the solution to the inequality is
 ,3   2,  .
We now need to sketch a graph of the solution set.
(c)
2
3
2
] (
0
3
or
3
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0
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Mat 117 Exam 1
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8. [6 pts] P dollars invested at interest rate r compounded annually increased to an
amount A given by
A  P(1  r ) 6
in 6 years. If an investment of $1000 is to increase to $1700 in 6 years, what is the
annual percentage rate for this investment? Show your work.
(Round your percent answer to two decimal places)
Solution:
To solve this problem we first must put the information we know into the equation.
1700  1000(1  r ) 6 . We now need to solve this equation. We will first divide by 1000
and then take the 6th root of both sides of the equation.
1700  1000(1  r ) 6
1700
 (1  r ) 6
1000
1.7  (1  r ) 6
6
1 .7  1  r
1 .7  1  r
.0924665633  r
Changing to a percent and rounding to 2 decimal places we find that the annual
percentage rate is 9.25%.
6
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Mat 117 Exam 1
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9. [8 pts] A subcompact car can be rented from Company A for $195 per week with no
extra charge for mileage. A similar car can be rented from Company B for $110 per
week , plus 25 cents for each mile driven. How many miles must you drive in a week
to make the rental fee for Company A more than that for Company B? Show your
work.
Solution:
This problem is asking us to find the range of mileage for one week for which the cost of
renting a car from Company A will be more than the cost of renting a car from Company
B. The cost of renting a car from Company A is a flat rate of $195. The cost of renting a
car from Company B is a flat rate of $110 plus a charge of 25 cents per mile driven. An
expression that can represent the cost of renting a car from Company B is 110  .25m
where m is the number of miles driven in a week. (Note that 25 cents was changed to
dollar form .25 since the flat rate was in dollars.) We can now write an inequality to help
us determine when the cost of renting from Company A is greater than the cost of renting
from Company B for a week. This inequality is 195  110  .25m . We will now solve
this inequality.
195  110  .25m
195  110  .25m
85  .25m
85
m
.25
340  m
We now can see that the number of miles driven must be less than 340 per week for
Company A to cost more than Company B.
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Mat 117 Exam 1
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10. [8 pts] An open box is to be made from a square piece of material by cutting 4-inch
squares from the corners and turning up the sides (see figure). The volume of the
finished box is to be 484 cubic inches. Find the original size of the square piece of
material. Show your work.
x
4
4
4
4
x
x
4
4
4
x
4
Solution:
For this problem we need to determine the volume of the box in terms of x that could be
make out of this square piece of material with 4- by 4-inch squares cut out of each corner
and then turning up the sides. The formula for volume of a box is
V  length  width  height . When the sides of the box are turned up the length of the
box is x, the width of the box is x, and the height of the box is 4. Putting this into the
equation, we find that the volume of the box in terms of x is V  x  x  4  4 x 2 . We
also know that the volume of the box is 484 cubic inches. This will allow us to write and
solve an equation to determine what the value of x is in the problem. The equation is
4 x 2  484
484
x2 
4
x 2  121
x   121
x  11
Solving the equation, we find that x  11. Since x represents a distance, we can discard
the negative number since distances cannot be negative. Thus x  11. We are not yet
done. The question asked us to find the original size of the square piece of material. x is
only the measurement of part of the side. We now need to add in the 4 inches that were
cut out of each side of the square. Thus the original material was 11  4  4  19 inches
on each side or was 361square inches.
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