By LearningExpress Editors
Updated on Sep 26, 2011
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The following explanations show one way in which each problem can be solved. You may have another method for solving these problems.
 b. Let l = the length of the rectangle and let w = the width of the rectangle. Since the width is 6 inches less than 3 times the length, one equation is w = 3l – 6. The formula for the perimeter of a rectangle is 2l + 2w = 104. Substituting the first equation into the perimeter equation for w results in 2l + 2(3l – 6) = 104. Use the distributive property on the left side of the equation: 2l + 6l – 12 = 104. Combine like terms on the left side of the equation: 8l – 12 = 104; add 12 to both sides of the equation: 8l – 12 + 12 = 104 + 12. Simplify to: 8l = 116. Divide both sides of the equation by 8: . Therefore, the length is l = 14.5 inches and the width is w = 3(14.5) – 6 = 37.5 inches.
 a. Let w = the width of the parallelogram and let l = the length of the parallelogram. Since the length is 5 more than the width, then l = w + 5. The formula for the perimeter of a parallelogram 2l + 2w = 50. Substituting the first equation into the second for l results in 2(w + 5) + 2w = 50. Use the distributive property on the left side of the equation: 2w + 10 + 2w = 50; combine like terms on the left side of the equation: 4w + 10 = 50. Subtract 10 on both sides of the equation: 4w + 10 – 10 = 50 – 10. Simply to: 4w = 40. Divide both sides of the equation by 4: ; w = 10. Therefore, the width is 10 cm and the length is 10 + 5 = 15 cm.
 c. Let x = the amount invested at 12% interest. Let y = the amount invested at 15% interest. Since the amount invested at 15% is 100 more then twice the amount at 12%, then y = 2x + 100. Since the total interest was $855, use the equation 0.12x + 0.15y = 855. You have two equations with two variables. Use the second equation 0.12x + 0.15y = 855 and substitute (2x + 100) for y: 0.12x + 0.15(2x + 100) = 855. Use the distributive property: 0.12x + 0.3x + 15 = 855. Combine like terms: 0.42x + 15 = 855. Subtract 15 from both sides: 0.42x + 15 – 15 = 855 – 15; simplify: 0.42x = 840. Divide both sides by 0.42: . Therefore, x = $2,000, which is the amount invested at 12% interest.
 c. Let x = the amount invested at 8% interest. Since the total interest is $405.50, use the equation 0.06(4,000) + 0.08x = 405.50. Simplify the multiplication: 240 + 0.08x = 405.50. Subtract 240 from both sides: 240 – 240 + 0.8x = 405.50 – 240; simplify: 0.08x = 165.50. Divide both sides by 0.08: . Therefore, x = $2,068.75, which is the amount invested at 8% interest.
 d. Let x = the amount of coffee at $3 per pound. Let y = the total amount of coffee purchased. If there are 18 pounds of coffee at $2.50 per pound, then the total amount of coffee can be expressed as y = x + 18. Use the equation 3x + 2.50(18) = 2.85y since the average cost of the y pounds of coffee is $2.85 per pound. To solve, substitute y = x + 18 into 3x + 2.50(18) = 2.85y. 3x + 2.50(18) = 2.85(x + 18). Multiply on the left side and use the distributive property on the right side: 3x + 45 = 2.85x + 51.30. Subtract 2.85x on both sides: 3x – 2.85x + 45 = 2.85x – 2.85x + 51.30. Simplify: 0.15x + 45 = 51.30. Subtract 45 from both sides: 0.15x + 45 – 45 = 51.30 – 45. Simplify: 0.15x = 6.30. Divide both sides by 0.15: ; so, x = 42 pounds, which is the amount of coffee that costs $3 per pound. Therefore, the total amount of coffee is 42 + 18, which is 60 pounds.
 c. Let x = the amount of candy at $1.90 per pound. Let y = the total number of pounds of candy purchased. If it is known that there are 40 pounds of candy at $2.15 per pound, then the total amount of candy can be expressed as y = x + 40. Use the equation 1.90x + 2.15(40) = $162 since the total amount of money spent was $162. Multiply on the left side: 1.90x + 86 = 162. Subtract 86 from both sides: 1.90x + 86 – 86 = 162 – 86. Simplify: 1.90x = 76. Divide both sides by 1.90: ; so, x = 40 pounds, which is the amount of candy that costs $1.90 per pound. Therefore, the total amount of candy is 40 + 40, which is 80 pounds.
 a. Let x = the amount of marigolds at $1 per packet. Let y = the amount of marigolds at $1.26 per packet. Since there are 50 more packets of the $1.26 seeds than the $1 seeds, y = x + 50. Use the equation 1x + 1.26y = 420 to find the total number of packets of each. By substituting into the second equation, you get 1x + 1.26(x + 50) = 402. Multiply on the left side using the distributive property: 1x + 1.26x + 63 = 402. Combine like terms on the left side: 2.26x + 63 = 402. Subtract 63 from both sides: 2.26x + 63 – 63 = 402 – 63. Simplify: 2.26x = 339. Divide both sides by 2.26: ; so, x = 150 packets, which is the number of packets that costs $1 each.
 a. Let x = the amount of 3% iodine solution. Let y = the amount of 20% iodine solution. Since the total amount of solution was 85 oz., then x + y = 85. The amount of each type of solution added together and set equal to the amount of 19% solution can be expressed in the equation 0.03x + 0.20y = 0.19(85); Use both equations to solve for x. Multiply the second equation by 100 to eliminate the decimal point: 3x + 20y = 19(85). Simplify that equation: 3x + 20y = 1805. Multiply the first equation by –20: –20x + –20y = –1700. Add the two equations to eliminate y: –17x + 0y = –85. Divide both sides of the equation by –17: ; x = 5. The amount of 3% iodine solution is 5 ounces.
 c. Let x = the amount of 34% acid solution. Let y = the amount of 18% iodine solution. Since the total amount of solution was 30 oz., then x + y = 30. The amount of each type of solution added together and set equal to the amount of 28% solution can be expressed in the equation 0.34x + 0.18y = 0.28(30). Use both equations to solve for x. Multiply the second equation by 100 to eliminate the decimal point: 34x + 18y = 28(30); simplify that equation: 34x + 22y = 840. Multiply the first equation by –18: –18x + –18y = –540. Add the two equations to eliminate y: 16x + 0y = 300. Divide both sides of the equation by 16: , x = 18.75. The amount of 34% acid solution is 18.75 ounces.
 b. Let x = Ellen's age and let y = Bob's age. Since Bob is 2 years from being twice as old as Ellen, than y = 2x – 2. The sum of twice Bob's age and three times Ellen's age is 66 and gives a second equation of 2y + 3x = 66. Substituting the first equation for y into the second equation results in 2(2x – 2) + 3x = 66. Use the distributive property on the left side of the equation: 4x – 4 + 3x = 66; combine like terms on the left side of the equation: 7x – 4 = 66. Add 4 to both sides of the equation: 7x – 4 + 4 = 66 + 4. Simplify: 7x = 70. Divide both sides of the equation by 7: . The variable, x, is now alone: x = 10. Therefore, Ellen is 10 years old.
 d. Let x = Shari's age and let y = Sam's age. Since Sam's age is 1 less than twice Shari's age this gives the equation y = 2x – 1. Since the sum of their ages is 104, this gives a second equation of x + y = 92. By substituting the first equation into the second for y, this results in the equation x + 2x – 1 = 92. Combine like terms on the left side of the equation: 3x – 1 = 92. Add 1 to both sides of the equation: 3x – 1 + 1 = 92 + 1. Simplify: 3x = 93. Divide both sides of the equation by 3: . The variable, x, is now alone: x = 31. Therefore, Shari's age is 31.
 d. Let x = the cost of one binder and let y = the cost of one pen. The first statement, "two binders and three pens cost $12.50," translates to the equation 2x + 3y = 12.50. The second statement, "three binders and five pens cost $19.50," translates to the equation:
 a. Let x = the number of degrees in the smaller angle and let y = the number of degrees in the larger angle. Since the angles are complementary, x + y = 90. In addition, since the larger angle is 15 more than twice the smaller, y = 2x + 15. Substitute the second equation into the first equation for y: x + 2x + 15 = 90. Combine like terms on the left side of the equation: 3x + 15 = 90. Subtract 15 from both sides of the equation: 3x + 15 – 15 = 90 – 15; simplify: 3x = 75. Divide both sides by 3: . The variable, x, is now alone: x = 25. The number of degrees in the smaller angle is 25.
 b. Let x = the cost of a student ticket. Let y = the cost of an adult ticket. The first sentence, "The cost of a student ticket is $1 more than half of an adult ticket," gives the equation x = + 1; the second sentence, "six adults and four student tickets cost $76," gives the equation 6y + 4x = 76. Substitute the first equation into the second for x: 6y + 4( + 1) = 76. Use the distributive property on the left side of the equation: 6y + 2y + 4 = 76. Combine like terms: 8y + 4 = 76. Subtract 4 on both sides of the equation: 8y + 4 – 4 = 76 – 4; simplify: 8y = 72. Divide both sides by 8: . The variable is now alone: y = 9. The cost of one adult ticket is $9.
 a. Let x = the cost of one shirt. Let y = the cost of one tie. The first part of the question, "three shirts and 5 ties cost $23," gives the equation 3x + 5y = 23; the second part of the question, "5 shirts and one tie cost $20," gives the equation 5x + 1y = 20. Multiply the second equation by –5: –25x – 5y = –100. Add the first equation to that result to eliminate y. The combined equation is: –22x = – 77. Divide both sides of the equation by –22: . The variable is now alone: x = 3.50; the cost of one shirt is $3.50.
 c. The terms 3x and 5x are like terms because they have exactly the same variable with the same exponent. Therefore, you just add the coefficients and keep the variable. 3x + 5x = 8x.
 c. Because the question asks for the difference between the areas, you need to subtract the expressions: 6a + 2 – 5a. Subtract like terms: 6a – 5a + 2 = 1a + 2; 1a = a, so the simplified answer is a + 2.
 b. Since the area of a rectangle is A = length times width, multiply (x^{3})(x^{4}). When multiplying like bases, add the exponents: x^{3}+4 = x^{7}.
 c. Since the area of the soccer field would be found by the formula A = length × width, multiply the dimensions together: 7y^{2} × 3xy. Use the commutative property to arrange like variables and the coefficients next to each other: 7 × 3 × x × y^{2} × y. Multiply: remember that y^{2} × y = y^{2} × y^{1} = y^{2} + 1 = y^{3}. The answer is 21xy^{3}.
 a. Since the area of a parallelogram is A = base times height, then the area divided by the base would give you the height; ; when dividing like bases, subtract the exponents; x^{8–4} = x^{4}.
Therefore, the cost of one pen is $1.50. Since the cost of 2 binders and 3 pens is 12.50, substitute y = 1.50 into the first equation: 3 × $1.50 = $4.50; $12.50 – 4.50 = $8.00; $8.00 ÷ 2= $4.00, so each binder is $4.00. The total cost of 1 binder and 1 pen is $4.00 + $1.50 = $5.50.
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
 Algebra Word Problems Practice Questions Set 4 (You are here)
 Algebra Word Problems Practice Questions Set 5
 Algebra Word Problems Practice Questions Set 6
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