Applications of Algebra in Word Problems Study Guide (page 3)
Introduction to Applications of Algebra in Word Problems
The human mind has never invented a labor-saving machine equal to algebra.
This lesson will cover some of the most common application questions involving algebra. Topics include consecutive integers, mixture problems, coin problems, age problems, and distance problems.
Consecutive Integer Problems
Consecutive integers are numbers in order one after the other. The integers 3, 4, and 5 are three consecutive integers. Because they are integers, there are no fractions or decimals.
- Consecutive odd integers are numbers such as 1, 3, 5, and –11, –9, –7.
- Consecutive even integers are numbers such as 4, 6, 8, and –20, –18, –16.
Since consecutive integers are always one apart from each other, use the expressions x, x + 1, x + 2, and so on, to define the variables in a consecutive integer problem.
Since consecutive odd integers are always two apart from each other, and consecutive even integers are always two apart as well, use the expressions x, x + 2, x + 4, and so on, to define the variables in a consecutive odd integer or consecutive even integer problem.
The sum of three consecutive integers is 60. What are the integers?
Read and understand the question. This question is looking for three consecutive integers that add to 60.
Make a plan. Use the expressions x, x + 1, and x + 2 to define the variables. Then, write an equation by adding these expressions and setting them equal to 60.
Carry out the plan. First, let x = the smallest integer, let x + 1 = the next integer, and let x + 2 = the greatest integer. Then, add these expressions together and set them equal to 60. The equation becomes x + x + 1 + x + 2 = 60. Combine like terms to get 3x + 3 = 60. Subtract 3 from each side of the equation.
- 3x + 3 – 3 = 60 – 3
- Simplify to get 3x = 57
- Divide each side by 3 to get the variable x alone.
- x = 19
- Because x = 19, then x + 1 = 20 and x + 2 = 21. The three integers are 19, 20, and 21.
- Check your answer. Check your answer by finding the sum of the three integers. The sum is 19 + 20 + 21 = 60, so this answer is checking.
Be sure to use your equation solving skills as outlined in Lesson 13 to help with solving word problems with equations. Refer back to the details in that lesson when necessary.
Twice the smaller of two consecutive even integers is equal to the larger even integer increased by 32. What are the two integers?
Read and understand the question. This question is looking for two consecutive even integers where clues are given for each.
Make a plan. Use the expressions x and x + 2 to define the variables. Then, write an equation by multiplying the smaller integer by 2 and setting this equal to the larger plus 32.
Carry out the plan. First, let x = the smaller even integer and let x + 2 = the larger even integer. Then, multiply the smaller by 2 to get 2x and increase the larger by 32 to get (x + 2) + 32. Set these values equal to each other. The equation becomes
- 2x = (x + 2) + 32
Combine like terms to get 2x = x + 34. Subtract x from each side of the equation.
- 2x – x = x – x + 34
Simplify to get x = 34. Because x = 34, then x + 2 = 36. The two integers are 34 and 36.
Check your answer. Check your answer by making sure the two integers fit the clues in the problem. Twice the smaller is 2 × 34 = 68, and the larger increased by 32 is 36 + 32 = 68. Because these values are equal, this answer is checking.
Mixture problems involve two or more different types of materials that will be combined to form one mixture. To solve mixture problems, use the clues in the problem to set up the let statements for the amounts of each type that will be mixed. Then, add the parts together and set the sum equal to the total mixture. Use the following example to help with these steps.
Jane wants to mix $4 per pound coffee with $6 per pound coffee to get a mixture that costs $5.50 per pound. If the total mixture contains 10 pounds, how many pounds of the $4 per pound coffee should she buy?
Read and understand the question. Jane is mixing together two different types of coffee. Each type is a different price. You are looking for the number of pounds of the less expensive coffee as a final answer.
Make a plan. Write an expression for the amounts of each type of coffee using the information given. Then, multiply each by the price per pound. Set this equal to the total mixture, which is equal to $5.50 × 10 or 5.50(10).
Carry out the plan. Let x = the number of pounds of the $4 per pound coffee. Since there is a total of 10 pounds in all, then 10 – x = the number of pounds of the $6 per pound coffee. Next, write an equation that adds the cost of the two types of coffee and sets it equal to the total. The cost of the $4 coffee is 4x, the cost of the $6 coffee is 6(10 – x), and the total cost is 5.50(10). The equation is
- 4x + 6(10 – x) = 5.50(10)
Simplify by using the distributive property:
- 4x + 60 – 6x = 55
Combine like terms:
- –2x + 60 = 55
Subtract 60 from each side of the equation:
- –2x + 60 – 60 = 55 – 60
Simplify to get –2x = –5. Divide each side by –2.
- x = 2.5
Because x represents the number of pounds of the $4 per pound coffee, Jane should buy 2.5 pounds.
Check your answer. To check this solution, first find the amount of the $6 per pound coffee. Using your solution, this should be 10 – 2.5 = 7.5 pounds. Next, multiply the number of pounds of each by the cost of each. $4 × 2.5 pounds is equal to $10 and $6 × 7.5 pounds is equal to $45. These amounts have a sum of $10 + $45 = $55, which was the cost of the total mixture. This answer is checking.
Coin problems are similar to mixture problems in that you are taking different values of the various coins and mixing them together to get a total amount. Take the following example.
Sebastian has a total of $2.75 in his bank. He has only quarters and dimes, and has 10 more dimes than quarters. How many of each coin does he have in his bank?
Read and understand the question. This question is looking for the total number of quarters and the total number of dimes in a bank containing only quarters and dimes. There are 10 more dimes than quarters.
Make a plan.Write an expression for the number of quarters and the number of dimes in the bank. Multiply each of these expressions by the value of each type of coin. Then, add these expressions and set them equal to the total of $2.75.
Carry out the plan. Let x = the number of quarters and let x + 10 = the number of dimes. A quarter has a value of $0.25, so multiply 0.25 by x to get 0.25x. A dime has a value of $0.10 so multiply 0.10 by x + 10 to get 0.10(x + 10). Next, write an equation by adding these expressions and setting them equal to $2.75. The equation is 0.25x + 0.10(x + 10) = 2.75. Use the distributive property to eliminate the parentheses. The equation becomes 0.25x + 0.10x + 1 = 2.75. Combine like terms to get 0.35x + 1 = 2.75. Subtract 1 from each side to get 0.35x = 1.75. Divide each side by 0.35.
- x = 5
Therefore, there are 5 quarters and 5 + 10 = 15 dimes.
Check your answer. Check this solution by making sure that the coins are equal to a total of $2.75. Five quarters has a value of $0.25 × 5 = $1.25, and 15 dimes has a value of $0.10 × 15 = $1.50. The sum is $1.25 + $1.50 = $2.75, so this answer is checking.
To solve age problems, use the clues within the problem to define a variable for each person in the problem. Then, write an equation based on the given information about their ages.
Tasha is 6 years older than Frank. If the sum of their ages is 54, how old is Tasha?
Read and understand the question. This question is looking for Tasha's age when clues are given about her age related to Frank's age.
Make a plan. Write an expression for each person's age, and add these expressions together for a total of 54.
Carry out the plan. Let x = Frank's age. Because Tasha is 6 years older, let x + 6 = Tasha's age. Add the ages together and set the sum equal to 54, since the sum of their ages is 54.
- x + x + 6 = 54
Combine like terms to get 2x + 6 = 54. Subtract 6 from each side to get
- 2x + 6 – 6 = 54 – 6 = 48
- 2x = 48
Divide each side of the equation by 2 to get x = 24. Therefore, Frank is 24 years old and Tasha is 24 + 6 = 30 years old.
Check your answer. To check this answer, add the ages and make sure they have a sum of 54 and that their difference is 6.
- 24 + 30 = 54
- 30 – 24 = 6
so this problem is checking.
D = R × T Problems (Distance = Rate × Time)
The formula distance = rate × time, or d = r × t, is commonly used in both math and science. This formula states that the rate, or the speed, at which something travels multiplied by the time spent traveling is equal to the distance traveled. Use this formula to solve the following problem.
Two people leave from the same city and drive in opposite directions. Person A is traveling at a rate of 55 miles per hour and person B is traveling at a rate of 60 miles per hour. If they leave the city at the exact same time, how long will it take for them to be 345 miles apart?
Read and understand the question. This question is looking for the time it will take two people leaving from the same point in opposite directions to be 345 miles apart. Each person is driving at a different rate.
Make a plan. Use the formula distance = rate × time, or d = r × t to solve this problem. Write an expression for the distance of each person, showing that the sum of these distances is 345 miles.
Carry out the plan. Each person's rate is given and the time is unknown, so use t for the time. Since distance = rate × time, person A's distance is 55t and person B's distance is 60t. Write an equation that adds these two distances and sets the sum equal to 345 miles.
- 55t + 60t = 345
Combine like terms to get 115t = 345. Divide each side of the equation by 115.
- t = 3
In three hours, they will be 345 miles apart.
Check your answer. To check this problem, substitute t = 3 for the time, and make sure that the total distance between them adds to 345 miles.
- 55 miles per hour × 3 hours = 165 miles
- 60 miles per hour × 3 hours = 180 miles
This is a sum of 165 + 180 = 345 miles, so this solution is checking.
Find practice problems and solutions for these concepts at Applications of Algebra in Word Problems Practice Questions.
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