College Algebra Worksheets
By O. Pauline Chow
HACC
Math 103 College Algebra
Lecture Notes for chapter 1.1-1.2 (audio dated 9/16/07 at 8:58 am, 60 minutes long)
Solving equations
Application problems
Equations in One Variable
When solving an equation which is not linear, you need
1) remove grouping symbols ( ), [ ], and { } using distributive property,
2) clear all denominators by multiplying the least common denominator, LCD, to each term in the equation and be sure to include the
restriction(s),
3) combine like terms on the same side,
3) group terms with the unknown on one side and terms without the unknown on the other side using addition property,
4) take the unknown out as a common factor, if possible, and
5) solve accordingly.
3x
x − 11
−5 =
x +1
x +1
The restriction is x ≠ –1. Multiply each term by the LCD, x + 1.
x − 11
3x
( x + 1)
− 5( x + 1) =
( x + 1)
x +1
x +1
3x − 5 x − 5 = x − 11
Ex 1
−2 x − 5 = x − 11
−3x − 5 = −11
−3x = −6
−6
=2
−3
Let us check the answer.
3(2)
(2) − 11
− 5?
=
(2) + 1
(2) + 1
6
−9
− 5?
= 3
3
True
−3 = −3
x=
The solution is {2}.
1
5
+
= 10.379
x 6.72
The restriction is x ≠ 0. The LCD is 6.72x.
1
5
6.72 x +
6.72 x = 10.379(6.72 x)
x 6.72
6.72 + 5 x = 69.74688 x
Ex 2
6.72 = 64.74688 x
x=
6.72
= 0.1037887849 ≈ 0.104
64.74688
Ex 3
0.25(2x – 1.6)2 = (x – 0.9)2
You need to foil each square first.
0.25(2x – 1.6)2 = (x – 0.9)2
0.25(2x – 1.6)(2x – 1.6) = (x – 0.9)(x – 0.9)
1
2
College Algebra Worksheets
By O. Pauline Chow
HACC
0.25(4x2 – 3.2x – 3.2x + 2.56) = x2 – 0.9x – 0.9x + 0.81
0.25(4x2 – 6.4x + 2.56) = x2 – 1.8x + 0.81
x2 – 1.6x + 0.64 = x2 – 1.8x + 0.81
0.2x = 0.17
x = 0.17/0.2 = 0.85
Ex 4
(–3.4 X 10–9)x + 3.45 X 10–8 = 1.63 X 104
3.45 X 10–8 is a number in scientific notation. To use the calculator, you need to use EE or EXP key to enter. Now solve the equation
by hand first and use the calculator at the end.
(–3.4 X 10–9)x + 3.45 X 10–8 = 1.63 X 104
(–3.4 X 10–9)x = 1.63 X 104 – 3.45 X 10–8
1.63X104 − 3.45X10−8
x=
−3.4X10−9
To enter into the calculator, be sure to use parentheses for the numerator:
(1.63 EE 4 – 3.45 EE –8) ÷ –3.4 EE –9 =
x ≈ –4.79411765 X 1012 ≈ –4.794 X 1012
Ex 5 9 – |2x – 3| = 6
9 – |2x – 3| = 6
–|2x – 3| = –3
|2x – 3| = 3
2x – 3 = –3 or 2x – 3 = 3
2x = 0 or 2x = 6
x = 0 or x = 3
isolate the absolute value term
|w| = a, a > 0 w = –a or w = a
3
4
7x − 3
−
=
x 1 − x x2 − x
First you need to factor the denominator, x2 – x and change 1 – x to –(x – 1).
3
4
7x − 3
x ≠ 0,1
−
=
x 1 − x x( x − 1)
3
4
7x − 3
−
=
x −( x − 1) x( x − 1)
3
4
7x − 3
x( x − 1) +
x( x − 1) =
x( x − 1)
x x −1
x( x − 1)
3( x − 1) + 4 x = 7 x − 3
3x − 3 + 4 x = 7 x − 3
7x − 3 = 7x − 3
Ex 6
Stop! This is an identity. So, the solution is the set of all real numbers, except 0 and 1 or in interval notation, (–∞, ∞), x ≠ 0, 1.
Constructing Models to Solve Problems
I select some application problems and show the steps and explanation.
Ex 1 Construction Penalties Gonzales Construction contracted Kentwood High and Memorial Stadium for a total cost of $4.7 million.
Because the construction was not completed on time, Gonzales paid 5% of the amount of the high school contract in penalties and 4% of
the amount of the stadium contract in penalties. If the total penalty was $223,000, then what was the amount of each contract?
3
College Algebra Worksheets
By O. Pauline Chow
HACC
To do this problem, we will use a table and label the columns and rows. Let us use column 1 represent all info about High School, column
2 about Stadium, and column 3 for the total.
High School
Stadium
Total
Next we will label each with the common info for each row. So row 1 represents % penalty, row 2 cost of each construction, and row 3
amount of penalty.
High School
5%
$x
$0.05x
% of penalty
Cost of construction
Amount of penalty
Stadium
4%
$(4700000 – x)
$0.04(4700000 – x)
Total
N/A
$4,700,000
$223,000
So, we know total amount of penalty is the sum of penalty from the high school and Stadium.
penalty from HS + penalty from Stadium = total penalty
0.05x + 0.04(4,700,000 – x) = 223,000
0.05x + 188,800 – 0.04x = 223,000
0.01x + 188,800 = 223,000
0.01x = 35,000
x = 3,500,000 = 3.5 million
4.7 – x = 4.7 – 3.5 = 1.2 million
Therefore, cost of construction for High School is $3.5 million and stadium is $1.2 million.
Ex 2 Mixing alcohol solutions A pharmacist needs to obtain a 70% alcohol solutions. How many ounces
of a 30% alcohol solution must be mixed with 40 ounces of an 80% alcohol solution to obtain a 70% alcohol
solution? [ans. 10 oz of 30% solution]
Solution I
% of alcohol
Solution II
Mixture
30%
80%
70%
Wt of solution
x oz
40 oz
(x + 40) oz
Wt of alcohol
0.30 x oz
0.80(40) oz
0.70(x + 40) oz
wt of alcohol from solution I + wt of alcohol from solution II = wt of alcohol from mixture
0.30x + 32 = 0.70(x + 40)
0.30x + 32 = 0.70x + 28
–0.4x = –4
x = 10
10 oz of 30 % solution.
Ex 3 Diluting bleach How much pure water must be added to a 4–liter solution that is 5% sodium hypochlorite (the main ingredient in
household bleach) to get a solution that is 3% sodium hypochlorite?
Remember, pure water does not have sodium hypochlorite.
4
College Algebra Worksheets
By O. Pauline Chow
HACC
% of sodium
hypochlorite
Vol of solution
Vol of sodium
hypochlorite
Pure water
Solution I
Mixture
0%
5%
3%
x liters
4 liters
(x + 4) liters
0x liters
0.05(4) liters
0.03(x + 4) liters
vol of SH from pure water + vol of SH from solution I = vol of SH from mixture
0 + 0.2 = 0.03(x + 4)
0.2 = 0.03x + 0.12
0.08 = 0.03x
x = 8/3 = 2 2/3
2 2/3 liters of pure water.
Ex 4 Increasing Area of a Field Julia's soybean field is 3 m longer than it is wide. To increase her
production, she plans to increase both the length and width by 2 m. If the new field is 46 m2 larger than the old
field, then what are the dimensions of the old field?
Let x m be the width. So the length is x + 3 m.
area of the old field is length times width = x(x + 3) m2
Now both length and width are increased by 2m. New width is x + 2 m and new length is x + 3 + 2 = x + 5 m.
New area = (x + 2)(x + 5) m2
Since new area is 46 m2 larger than the old field, we have
(x + 2)(x + 5) = 46 + x(x + 3)
x2 + 7x + 10 = 46 + x2 + 3x
4x = 36, x = 9
So width is 9 m and length is 9 + 3 = 12 m.
Ex 5 Fencing dog pens Clint is constructing two adjacent rectangular dog pens. Each pen will be three
times as long as it is wide, and the pens will share a common long side. If Clint has 65 ft of fencing, what are
the dimensions of each pen? (Perimeter of a figure = sum of lengths of all sides)
w
3w
w
3w
w
3w
w
Let w be the width of each pen. Since the long side is 3 times its width, length is 3w. Now you just need to
add all the sides and find the total amount of fencing.
3w + 3w + 3w + w +w + w + w = 65
13w = 65
w=5
So the dimensions are 5 ft by 15 ft for each pen.
5
College Algebra Worksheets
By O. Pauline Chow
HACC
Ex 6 Processing Forms Rita can process a batch of insurance claims in 4 hr working alone. Ed can
process a batch of insurance claims in 2 hr working alone. How long would it take them to process a batch of
claims if they worked together?
For this type of problem, you always need to consider the rate, time, and work completed.
Rate is how much work Rita can do in 1 hour. So, if Rita can process all forms in 4 hours, then her rate is ¼.
The rate for Ed is ½. Working together, if it takes Rita and Ed x hours to process all forms, then we
can find the work completed by each.
Method I: If we interpret rate as fraction of job done per hour, then we can set up the equation as:
Fraction of job done by Rita + Fraction of job done by Ed = Fraction of job done by both
1 1 1
+ =
4 2 x
3 1
=
4 x
4
1
x = =1
3
3
So, it will take both 1 1/3 hours to process all forms.
Method II: If we interpret rate as how fast each person can do in 1 hour, then we set up the equation for worked completed.
Rate
Rita
time on the job
Work completed
1/4
x hours
1/4x
Ed
½
x hours
1/2x
Total
N/A
N/A
1
1
1
x + x =1
4
2
x + 2x = 4
3x = 4
4
1
x = = 1 hours
3
3
Ex 7 How many gallons of 90% antifreeze solution must be mixed with 70 gallons of 20% antifreeze to have a mixture that is 80%
antifreeze?
Solution I
% of antifreeze
Solution II
Mixture
90%
20%
80%
Wt of solution
x gallons
70 gallons
(x + 70) gallons
Wt of antifreeze
0.90 x gallons
0.20(70) gallons
0.80(x + 70) gallons
wt of antifreeze in solution I + wt of antifreeze in solution II = wt of antifreeze in mixture
0.90x + 14 = 0.80(x + 70)
0.90x + 14 = 0.80x + 56
0.1x = 42 , x = 420
420 gallons of 90 % solution.
6
College Algebra Worksheets
By O. Pauline Chow
HACC
Ex 8 Dried bananas sell for $0.80 per quarter pound, and dried apricots sell for $1.00 per quarter pound. How many pounds of
apricots should be mixed with 6 pounds of bananas to get a mixture that sells for $0.95 per quarter pound? How many pounds of
apricots are needed?
Since there are 4 quarters in a pound, the cost of bananas per pound is $0.80(4) = $3.20 per pound.
Since the apricots sell for $1.00 per quarter pound, the cost is $4.00 per pound.
If the cost of mixture is 0.95 per quarter pound, then the cost is $3.80 per pound.
Cost per pound
Weight of each
Total cost of each
Bananas
$3.20
6 pounds
$3.20(6)
$19.2
apricots
$4.00
x pounds
$4.00x
4x
mixture
$3.80
(x + 6) pounds
$3.80(x + 6)
Total cost of bananas + total cost of apricots = total cost of mixture
19.2 + 4x = 3.80(x + 6)
19.2 + 4x = 3.80x + 22.8
19.2 + 0.20x = 22.8
0.2x = 3.6 x = 18
You need to mix 18 pounds of apricots with 6 pounds of bananas in order to set the selling price of the mixture at $0.95 per quarter
pound.
1
2
of the midterm exam score plus of the final exam score. If a student scored a 53 on
3
3
his midterm, then for what range of score on the final exam would he get a grade of C or better? A 70 is the lowest average to get a C.
The range of score is [
,100].
Ex 9 A student's grade is computed by taking
First you need to set up an unknown for the final exam score, say, x.
1
2
of midterm + of final exam is at least 70
3
3
1
2
(53) + x ≥ 70
3
3
53 + 2 x ≥ 210
2 x ≥ 157
x ≥ 78.5
The range of score is [78.5, 100] on the final exam.
Grade =
Note: If the student’s grade is computed by the average of the midterm and final exam scores, what would be the range? In this case,
you can see that he has to earn much higher in his final exam in order to bring the grade up.
Average grade is the sum of all test scores divided by the number of tests.
x + 53
≥ 70
2
x + 53 ≥ 140
x ≥ 87
He needs to score at 87 in order to get at least a C. The range is [87, 100].
Ex 10 A ship has already traveled 1057 nautical miles at an average speed 19 nautical miles per hour (knots). A second ship leaves
port to join in a race. If the second ship continues at a current pace, how fast must the second ship travel in order to catch the first ship
by the end of the 3926 nautical mile race?
The second ship must average
knots in order to catch the first ship by the end of the race.
The first ship has traveled 1057 nautical miles when the second ship starts at the same port.
College Algebra Worksheets
By O. Pauline Chow
HACC
Since the second ship will catch up with the first ship at the end of 3926 nautical miles, the time taken by the first ship to travel
(3926 – 1057) = 2869 miles is the same as that by the second ship to travel 3926 miles.
Time = distance/speed
Let x nautical miles per hour be the speed of the second ship.
Time taken by the first ship = (3926-1057)/19 = 2869/19 = 151 hours
Time taken by the second ship = 3926/x hours
3926
= 151
x
151x = 3926
x=
3926
= 26
151
The second ship was going at 26 nautical miles per hour.
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