Solving Equations with Variables on Both Sides – Blue Problems
1. Tyrone announces, “I just found $5.00. I now have five times more money
than if I had lost $5.00.” How many dollars did Tyrone have before finding
the $5.00? Express your answer to the nearest hundredth.
2. To find the weight, in pounds, of the largest gold nugget ever found, you can
solve this puzzle. If you take the number of pounds, divide it by 2 and then
subtract 8, you get the same value as if you triple the number of pounds and
then subtract 398. How many pounds does the nugget weigh?
3. Car Rental Quandary. I need to rent a car for my upcoming trip. Rent-AGem charges $20.25 per day plus 14 cent per mile. Super Saver Rentals
charges $18.25 per day plus 22 cent per mile.
At first glance it looks as if I should go with Super Saver Rentals. Still,
I’m concerned about the per-mile charge because I plan to do a lot of
driving during all three days of my trip.
How many miles would I have to drive to make the cost of renting a car
from Rent-A-Gem the same as the cost of renting a car from Super Saver
Rentals?
4. Duke is Missing. When Megan arrived home from school she found that
someone, she suspected her mean brother Bobby, had left the yard gate
open and her dog Duke had disappeared.
She immediately called the local S.P.C.A., and the man on the phone asked
her to describe Duke. Megan said, “He is a black-and-white spotted English
Setter. His left eye is black and his right eye is white.”
At the moment Bobby grabbed the phone from her hand and said, “Duke’s
head is 6 inches long, his tail is as long as his head plus half the length of
his body, and his body is as long as the head and tail combined.”
How long is Duke?
For exercises 5 – 9, use the following information. Check your answer with a
table.
Two grocery stores sell rice in bulk. The first charges $0.55 per pound. The second
charges $0.75 per pound for up to 3 pounds and $0.40 per pound for anything over
3 pounds.
5. Write expressions for the cost of rice at each store in terms of the
number of pounds bought, assuming you buy more than 3 pounds.
6. Write and solve an equation that relates the two expressions from exercise
5.
7. Interpret your result from exercise 6.
8. Evaluate each expression from Exercise 5 for the value of x from exercise
6. Interpret the result.
9. Describe what happens for values of x less than the one found in Exercise 6
and for values greater than it.
Distance, Rate, Time Problems
10. A train a mile and a half long takes a minute and a half to go through a
tunnel a mile and a half long. How fast is it going?
11. At 9 a.m. a car (A) began a journey from a point, traveling at 40 mph. At 10
a.m. another car (B) started traveling from the same point at 60 mph in the
same direction as car (A). At what time will car B pass car A?
12. By car, John traveled from city A to city B in 3 hours. At a rate that was
20 mph higher than John’s, Peter traveled the same distance in 2 hours.
Find the distance between the two cities.
13. Two trains started from the same point. At 8:00 a.m. the first train
traveled East at the rate of 80 mph. At 9:00 a.m., the second train traveled
West at the rate of 100 mph. At what time were they 530 miles apart?
14. Gary started driving at 9:00 a.m. from city A towards city B at a rate of 50
mph. At a rate that is 15 mph higher than Gary’s, Thomas started driving at
the same time as Gary from city B towards city A. If Gary and Thomas
crossed each other at 11 a.m., what is the distance between the two cities?
Solving Equations with Variables on Both Sides – Blue Solutions
1.
Translating Tyrone’s words to algebra, we get the following equation:
Distributing, we have
started with $7.50.
2.
3.
x + 5 = 5(x - 5).
x + 5 = 5 x - 25, which leads to 30 = 4 x and x = 75. Tyrone must have
⎛ x⎞
⎜ ⎟ - 8 = 3 x - 398. Adding 398 to
⎝2⎠
both sides and then multiplying both sides by 2, we get x + 780 = 6 x. Subtracting x from
each side, we get 780 = 5 x. Finally, dividing 780 by 5, we get x = 156, which means that the
Translating the English to algebra, we get the equation
nugget weighed 156 pounds.
Car Rental Quandary.
First, we set up an equation.
20.25x + .14y = 18.25x + .22y
The x = days of the trip, and the y = miles. Since his trip is 3 days, we can substitute the 3
for the x. Now, the equation is
60.75 + .14y = 54.75 + .22y
Then we subtracted .14y from each side. That makes the equation
60.75 = 54.75 + .08y
To get just one number and one variable on opposite sides, I subtract the 54.75 from 60.75.
That makes the equation
6 = .08y
To solve for y, you have to divide each side by .08.
Now you have solved the equation. In the three day span, he has to drive 75 miles.
4.
Duke is Missing.
The following was given as information with “b” being used to represent the length of the
body:
head
6 inches
tail
6 + .5 b
body
6 + 6 + .5b or b
The resulting equation is
6 + 6 + .5b = b
12 = .5b
24 = b
Therefore substituting into the original formulas you determine that:
head
6 inches
tail
6 + .5b or 6 + .5(24) or 18 inches
body
6 + 6 + .5b or 6 + 6 + .5(24) or 24 inches
Total length is 6 + 18 + 24 = 48 inches long.
5.
0.55 x, (0.75)3 + 0.4(x - 3)
6.
0.55
7.
7 lb of rice costs the same at both stores.
8.
3.85; 7 lb of rice costs $3.85 at both stores.
x = (0.75)3 + 0.4(x - 3); 7
9.
Rice is cheaper at the first store for less than 7 pounds and at the second store for more
than 7 pounds.
Distance, Rate, Time Problems
10.
The train is going 120 miles per hour (mph).
11.
After t hours the distances D1traveled by car A is given by
D1 = 40t
Car B starts at 10 a.m. and will therefore have spent one hour less than car when it passes it.
After (t – 1) hours, distance D2 traveled by car B is given by
D2 = 60(t – 1)
When car B passes car A, they are at the same distance from the starting point and
therefore D1 = D2 which gives
Solve the above equation for t to find
Car B passes car A at
12.
40t = 60(t – 1)
t = 3 hours
9 + 3 = 12 p.m.
Let x be John’s rate in traveling between the two cities. The rate of Peter will be x + 10. We
use the rate-time-distance formula to write the distance D traveled by John and Peter
(same distance D)
D = 3x and D = 2(x + 20)
The first equation can be solved for x to give
x=
Substitute x by
D into the second equation
3
D=2(
D
3
D + 20)
3
Solve for D to obtain
13.
D = 120 miles
When the first train has traveled for t hours the second train will have traveled (t – 1) hours
since it started 1 hour late. Hence if D1 and D2 are the distances traveled by the two trains,
then
D1 = 80t and D2 = 100(t – 1)
Since the trains are traveling in the opposite directions, the total distance D between the
two trains is given by
D = D1 + D2 = 180t – 100
14.
For D to be 530 miles, we need to have
180t – 100 = 530
Solve for t
t = 3 hours 30 minutes.
8 a.m. + 3:30 = 11:30 a.m.
Let D be the distance between the two cities. When Gary and Thomas cross each other, they
have covered all the distance between the two cities. Hence
D1 = 2 x 50 = 100 miles, distance traveled by Gary
D1 = 2 x (50 + 15) = 130 miles, distance traveled by Gary
Distance D between the two cities is given by
D = 100 miles + 130 miles = 230 miles
Bibliography Information
Teachers attempted to cite the sources for the problems included in this problem set. In some
cases, sources were not known.
Problems
Bibliography Information
3 – 4, 10
The Math Forum @ Drexel
(http://mathforum.org/)
1-2
Math Counts (http://mathcounts.org)
5-9
Larson, Ron, Laurie Boswell, Timothy D.
Kanold, and Lee Stiff. Algebra 1
Concepts and Skills. Evanston: McDougal
Littell, 2001. Print.