Solving Equations with Variables on Both Sides Blue Problems

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-A- Gem 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: x + 5 = 5(x - 5). Distributing, we have x + 5 = 5 x - 25, which leads to 30 = 4 x and x = 75. Tyrone must have started with $7.50. 2. 骣x Translating the English to algebra, we get the equation 琪 - 8 = 3 x - 398. Adding 398 to 桫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 nugget weighed 156 pounds. 3. 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 x = (0.75)3 + 0.4(x - 3); 7

7. 7 lb of rice costs the same at both stores.

8. 3.85; 7 lb of rice costs $3.85 at both stores. 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 40t = 60(t – 1) Solve the above equation for t to find t = 3 hours Car B passes car A at 9 + 3 = 12 p.m. 12. 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 D x = 3 D Substitute x by 3 into the second equation D D = 2 ( 3 + 20) Solve for D to obtain D = 120 miles 13. 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 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. 14. 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

The Math Forum @ Drexel 3 – 4, 10 (http://mathforum.org/)

1 - 2 Math Counts (http://mathcounts.org)

Larson, Ron, Laurie Boswell, Timothy D. Kanold, and Lee Stiff. Algebra 1 5 - 9 Concepts and Skills. Evanston: McDougal Littell, 2001. Print.