Example of algebra word problems are numerous. The goal of this unit is to give you the skills that you need to solve a variety of these algebra word problems. Show Example #1: A football team lost 5 yards and then gained 9. What is the team's progress? Solution For lost, use negative. For gain, use positive. Progress = -5 + 9 = 4 yards Example #2: Use distributive property to solve the problem below: Maria bought 10 notebooks and 5 pens costing 2 dollars each.How much did Maria pay? Solution 2 × (10 + 5) = 2 × 10 + 2 × 5 = 20 + 10 = 30 dollars Example #3: A customer pays 50 dollars for a coffee maker after a discount of 20 dollars What is the original price of the coffee maker? Solution Let x be the original price. x - 20 = 50 x - 20 + 20 = 50 + 20 x + 0 = 70 x = 70 Example #4: Half a number plus 5 is 11.What is the number? Solution Let x be the number. Always replace "is" with an equal sign (1/2)x + 5 = 11 (1/2)x + 5 - 5 = 11 - 5 (1/2)x = 6 2 × (1/2)x = 6 × 2 x = 12 Example #5: The sum of two consecutive even integers is 26. What are the two numbers? Solution Let 2n be the first even integer and let 2n + 2 be the second integer 2n + 2n + 2 = 26 4n + 2 = 26 4n + 2 - 2 = 26 - 2 4n = 24 n = 6 So the first even integer is 2n = 2 × 6 = 12 and the second is 12 + 2 = 14 Below are more complicated algebra word problems Example #6: The ratio of two numbers is 5 to 1. The sum is 18. What are the two numbers? Solution Let x be the first number. Let y be the second number x / y = 5 / 1 x + y = 18 Using x / y = 5 / 1, we get x = 5y after doing cross multiplication Replacing x = 5y into x + y = 18, we get 5y + y = 18 6y = 18 y = 3 x = 5y = 5 × 3 = 15 As you can see, 15/3 = 5, so ratio is correct and 3 + 15 = 18, so the sum is correct. Example #7: Algebra word problems can be as complicated as example #7. Study it carefully! Peter has six times as many dimes as quarters in her piggy bank. She has 21 coins in her piggy bank totaling $2.55 How many of each type of coin does she have? Solution Let x be the number of quarters. Let 6x be the number of dimes Since one quarter equals 25 cents, x quarters equals x × 25 cents or 25x cents Since one dime equals 10 cents, 6x dimes equals 6x × 10 cents or 60x cents Since one 1 dollar equals 100 cents, 2.55 dollars equals 2.55 × 100 = 255 cents Putting it all together, 25x cents + 60x cents = 255 cents 85x cents = 255 cents 85x cents / 85 cents = 255 cents / 85 cents x = 3 6x = 6 × 3 = 18 Therefore Peter has 3 quarters and 18 dimes Example #8: The area of a rectangle is x2 + 4x -12. What are the dimensions of the rectangle (length and width)? Solution The main idea is to factor x2 + 4x -12 Since -12 = -2 × 6 and -2 + 6 = 4 x2 + 4x -12 = ( x + -2) × ( x + 6) Since the length is usually longer, length = x + 6 and width = x + -2 Example #9: A must know how when solving algebra word problems The area of a rectangle is 24 cm2. The width is two less than the length. What is the length and width of the rectangle? Solution Let x be the length and let x - 2 be the width Area = length × width = x × ( x - 2) = 24 x × ( x - 2) = 24 x2 + -2x = 24 x2 + -2x - 24 = 0 Since -24 = 4 × -6 and 4 + -6 = -2, we get: (x + 4) × ( x + -6) = 0 This leads to two equations to solve: x + 4 = 0 and x + -6 = 0 x + 4 = 0 gives x = -4. Reject this value since a dimension cannot be negative x + -6 = 0 gives x = 6 Therefore, length = 6 and width = x - 2 = 6 - 2 = 4 Example #10: The sum of two numbers is 16. The difference is 4. What are the two numbers? Let x be the first number. Let y be the second number x + y = 16 x - y = 4 Solution Let x be the first number. Let y be the second number x + y = 16 x - y = 4 Solve the system of equations by elimination Adding the left sides and the right sides gives: x + x + y + -y = 16 + 4 2x = 20 x = 10 Since x + y = 16, 10 + y = 16 10 + y = 16 10 - 10 + y = 16 - 10 y = 6 The numbers are 10 and 6 The algebra word problems I solved above are typical questions. You will encounter them a lot in algebra. Hope you had fun solving these algebra word problems. Other algebra word problems/related topicsHave Algebra Word ProblemsShare it here with a solution! What Other Visitors Have SaidClick below to see contributions from other visitors to this page... Bessy and Bob algebra word problem Bessy has 6 times as much money as Bob. But when each earns $6, Bessy will have 3 times as much money as Bob. 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How do you solve an algebraic expression word problem?To solve an algebraic word problem:. Define a variable.. Write an equation using the variable.. Solve the equation.. If the variable is not the answer to the word problem, use the variable to calculate the answer.. What is an example of an algebra problem?Below are some of the examples of algebraic expressions.
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For example.. What are 2 examples of algebraic expression?An algebraic expression can be a combination of both variables and constants.. (a + b)2 = a2 + 2ab + b. ... . (a – b)2 = a2 – 2ab + b. ... . a2 – b2 = (a – b)(a + b). (a + b)3 = a3 + b3 + 3ab(a + b). (a – b)3 = a3 – b3 – 3ab(a – b). a3 – b3 = (a – b)(a2 + ab + b2). a3 + b3 = (a + b)(a2 – ab + b2). |