By LearningExpress Editors
Updated on Sep 26, 2011
Answers
The following explanations show one way in which each problem can be solved. You may have another method for solving these problems.
 a. Two consecutive odd integers are numbers in order like 3 and 5 or –31 and –29, which are each 2 numbers apart. In this problem you are looking for 2 consecutive odd integers. Let x = the first and smallest consecutive odd integer. Let x + 2 = the second (and larger) consecutive negative odd integer. Sum is a key word for addition so the equation becomes (x)+ (x + 2) = –112. Combine like terms on the left side of the equation: 2x + 2 = –112. Subtract 2 from both sides of the equation: 2x + 2 – 2 = –112 – 2; simplify: 2x = –114. Divide each side of the equation by 2: . The variable is now alone: x = –57. Therefore the larger value is: x + 2 = –55.
 c. Three consecutive even integers are numbers in order like 4, 6, and 8 or –30, –28 and –26, which are each 2 numbers apart. Let x = the first and smallest consecutive even integer. Let x + 2 = the second consecutive even integer. Let x + 4 = the third and largest consecutive even integer. Sum is a key word for addition so the equation becomes (x)+ (x + 2) + (x + 4) = 114. Combine like terms on the left side of the equation: 3x + 6 = 114. Subtract 6 from both sides of the equation: 3x + 6 – 6 = 114 – 6; simplify: 3x = 108. Divide each side of the equation by 3: . The variable is now alone: x = 36; therefore the next larger integer is: x + 2 = 38. The largest even integer would be: x + 4 = 40.
 d. Let t = the amount of time traveled. Using the formula distance = rate × time, substitute the rates of each car and multiply by t to find the distance traveled by each car. Therefore, 63t = distance traveled by one car and 59t = distance traveled by the other car. Since the cars are traveling in opposite directions, the total distance traveled by both cars is the sum of these distances: 63t + 59t. Set this equal to the total distance of 610 miles: 63t + 59t = 610. Combine like terms on the left side of the equation: 122t = 610. Divide each side of the equation by 122: the variable is now alone: t = 5. In 5 hours, the cars will be 610 miles apart.
 d. Use the formula distance = rate × time for each train and add these values together so that the distance equals 822 miles. For the first train, d = 65t and for the second train d = 72t, where d is the distance and t is the time in hours. Add the distances and set them equal to 822: 65t + 72t = 822. Combine like terms on the left side of the equation: 137t = 822; divide both sides of the equation by 137: . The variable is now alone: t = 6. In 6 hours, they will be 822 miles apart.
 d. Use the formula distance = rate × time for each train and add these values together so that the distance equals 1,029 miles. For the first train, d = 45t and for the second train d = 53t, where d is the distance and t is the time in hours. Add the distances and set them equal to 1,029: 45t + 53t = 1,029. Combine like terms on the left side of the equation: 98t = 1,029; divide both sides of the equation by 98: . The variable is now alone: t = 10.5 hours. The two trains will pass in 10.5 hours.
 c. Translate the sentence, "Nine minus five times a number is no less than 39," into symbols: 9 – 5x ≥ 39. Subtract 9 from both sides of the inequality: 9 – 9 – 5x ≥ 39 – 9. Simplify: –5x ≥ 30; divide both sides of the inequality by –5. Remember that when dividing or multiplying each side of an inequality by a negative number, the inequality symbol changes direction: . The variable is now alone: x ≤ –6.
 a. This problem is an example of a compound inequality, where there is more than one inequality in the question. In order to solve it, let x = the total amount of gumdrops Will has. Set up the compound inequality, and then solve it as two separate inequalities. Therefore, the second sentence in the problem can be written as: 2 < x – 2 < 6. The two inequalities are: 2 < x – 2 and x – 2 < 6. Add 2 to both sides of both inequalities: 2 + 2 < x – 2 + 2 and x – 2 + 2 < 6 + 2; simplify: 4 < x and x < 8. If x is greater than four and less than eight, it means that the solution is between 4 and 8. This can be shortened to: 4 < x < 8.
 a. This inequality shows a solution set where y is greater than or equal to 3 and less than or equal to eight. Both –3 and 8 are in the solution set because of the word inclusive, which includes them. The only choice that shows values between –3 and 8 and also includes them is choice a.
 b. Let x = the number. Remember that quotient is a key word for division, and at least means greater than or equal to. From the question, the sentence would translate to: + 5 ≥ x. Subtract 5 from both sides of the inequality: + 5 – 5 ≥ x – 5; simplify: ≥ x – 5. Multiply both sides of the inequality by 2: × 2 ≥ (x – 5) × 2; simplify: x ≥ (x – 5)2. Use the distributive property on the right side of the inequality: x ≥ 2x – 10. Add 10 to both sides of the inequality: x + 10 ≥ 2x – 10 + 10; simplify: x + 10 ≥ 2x. Subtract x from both sides of the inequality: x – x + 10 ≥ 2x – x. The variable is now alone: 10 ≥ x. The number is at most 10.
 d. Let x = the amount of hours Cindy worked. Let 2x + 3 = the amount of hours Carl worked. Since the total hours added together was at most 48, the inequality would be (x) + (2x + 3) ≤ 48. Combine like terms on the left side of the inequality: 3x + 3 ≤ 48. Subtract 3 from both sides of the inequality: 3x + 3 – 3 ≤ 48 – 3; simplify: 3x ≤ 45. Divide both sides of the inequality by 3: ; the variable is now alone: x ≤ 15. The maximum amount of hours Cindy worked was 15.
 b. Choices a and d should be omitted because the negative values should not make sense for this problem using time and cost. Choice b substituted would be 6 = 2(2) + 2 which simplifies to 6 = 4 + 2. Thus, 6 = 6. The coordinates in choice c are reversed from choice b and will not work if substituted for x and y.
 a. Let x = the total minutes of the call. Therefore, x – 1 = the additional minutes of the call. This choice is correct because in order to calculate the cost, the charge is 35 cents plus 15 cents times the number of additional minutes. If y represents the total cost, then y equals 0.35 plus 0.15 times the quantity x – 1. This translates to y = 0.35 + 0.15(x – 1) or y = 0.15(x – 1) + 0.35.
 d. Let x = the total miles of the ride. Therefore, x – 1 = the additional miles of the ride. The correct equation takes $1.25 and adds it to $1.15 times the number of additional miles, x – 1. Translating, this becomes y (the total cost) = 1.25 + 1.15(x – 1), which is the same equation as y = 1.15(x – 1) + 1.25.
 c. The total amount will be $4.85 plus two times the number of ounces, x. This translates to 4.85 + 2x, which is the same as 2x + 4.85. This value needs to be less than or equal to $10, which can be written as 2x + 4.85 ≤ 10.
 b. Let x = the number of checks written that month. Green Bank's fees would therefore be represented by .10x + 3 and SavingsRUs would be represented by .05x + 4.50. To find the value for which the banks charge the same amount, set the two expressions equal to each other: .10x + 3 = .05x + 4.50. Subtract 3 from both sides: .10x + 3 – 3 = .05x + 4.50 – 3. This now becomes: .10x = .05x + 1.50. Subtract .05x from both sides of the equation: .10x – .05x = .05x – .05x + 1.50; this simplifies to: .05x = 1.50. Divide both sides of the equation by .05: . The variable is now alone: x = 30. Costs would be the same if 30 checks were written.
 d. Let x = the number of miles traveled in the taxi. The expression for the cost of a ride with Easy Rider would be 1.25x + 2. The expression for the cost of a ride with Luxury Limo is 1x + 3.25. To solve, set the two expressions equal to each other: 1.25x + 2 = 1x + 3.25. Subtract 2 from both sides: 1.25x + 2 – 2 = 1x + 3.25 – 2. This simplifies to: 1.25x = 1x + 1.25; subtract 1x from both sides: 1.25x – 1x = 1x – 1x + 1.25. Divide both sides of the equation by .25: . The variable is now alone: x = 5; the cost would be the same if the trip were 5 miles long.
 b. Let x = the first integer and let y = the second integer. The equation for the sum of the two integers is x + y = 36, and the equation for the difference between the two integers is x – y = 6. To solve these by the elimination method, combine like terms vertically and the variable of y cancels out.
 c. Let x = the greater integer and y = the lesser integer. From the first sentence in the question we get the equation x = y + 2. From the second sentence in the question we get y + 2x = 7. Substitute x = y + 2 into the second equation: y + 2(y + 2) = 7; use the distributive property to simplify to: y + 2y + 4 = 7. Combine like terms to get: 3y + 4 = 7; subtract 4 from both sides of the equation: 3y + 4 – 4 = 7 – 4. Simplify to 3y = 3. Divide both sides of the equation by 3: ; therefore y = 1. Since the greater is two more than the lesser, the greater is 1 + 2 = 3.
 d. Let x = the lesser integer and let y = the greater integer. The first sentence in the question gives the equation y = 4x. The second sentence gives the equation x + y = 65. Substitute y = 4x into the second equation: x + 4x = 65. Combine like terms on the left side of the equation: 5x = 65; divide both sides of the equation by 5: . This gives a solution of x = 13, which is the lesser integer.
 a. Let x = the lesser integer and let y = the greater integer. The first sentence in the question gives the equation 3y + 5x = 9. The second sentence gives the equation y – 3 = x. Substitute y – 3 for x in the second equation: 3y + 5(y – 3) = 9. Use the distributive property on the left side of the equation: 3y + 5y – 15 = 9. Combine like terms on the left side: 8y – 15 = 9; add 15 to both sides of the equation: 8y – 15 + 15 = 9 + 15. Simplify to: 8y = 24. Divide both sides of the equation by 8: . This gives a solution of y = 3. Therefore the lesser, x, is three less than y, so x = 0.
This results in: , so x = 21
Substitute the value of x into the first equation to get 21 + y = 36. Subtract 21 from both sides of this equation to get an answer of y = 15.
More practice problems on algebra word problems can be found at:
 Algebra Word Problems Practice Questions Set 1
 Algebra Word Problems Practice Questions Set 2
 Algebra Word Problems Practice Questions Set 3 (You are here)
 Algebra Word Problems Practice Questions Set 4
 Algebra Word Problems Practice Questions Set 5
 Algebra Word Problems Practice Questions Set 6
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