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# Word Problems in Algebra

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Steps to Solving word problems

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### Word Problems in Algebra

1. 1. Word Problems in Algebra<br />Word problems allow us to apply algebra to real life situations. Many students find them difficult, but if approached in a logical manner, they can be easily solved. <br />
3. 3. Some Basic Examples<br />Three less than twice a number is 13. What is the number?<br />FIND:we need to find the number, use the variable: x<br />FACTS: twice the number is 2x, three less than that is 2x -3<br />FORMULA: 2X – 3 = 13<br />SOLVE: 2x - 3 = 13<br /> 2x – 3 + 3 = 13 + 3<br /> 2x = 16<br /> X = 8<br />ANSWER: Does the answer 8 make sense? Twice 8 is equal to 16. Three less than 16 is 13. YES it answers the original problem and makes sense.<br /><ul><li>add 3 to both sides (addition property of equality)
4. 4. divide both sides by 2 (division property of equality)</li></li></ul><li>Consecutive Integer problems<br />Twice the sum of three consecutive odd integers is 150. Find the three numbers.<br />FIND: the three numbers; we’ll use the variable x for one of them. Each consecutive odd integer is two more than the previous, so we will represent the 3 integers as : x, x+2, x+4 <br />FACTS: twice the sum = 150, so we need to represent the sum of the integers, and twice that value. Sum of integers will be x + x+2 + x + 4 and twice that value will be 2(x + x+2 + x + 4)<br />FORMULA: 2(x + x +2 + x + 4) = 150<br />SOLVE: 2(x + x +2 + x + 4) = 150<br /> 2(3x + 6) = 150<br /> 6x + 12 = 150<br /> 6x + 12 – 12 = 150 - 12<br /> 6x = 138<br /> x = 23 <br />ANSWER: We have solved for x, but we need to supply three answers: x = 23, x + 2 = 25, x + 4 = 27. Check: the sum of the integers is 75; twice that is 150<br /><ul><li>Combine like terms
5. 5. Distribute
6. 6. Subtract 12 from both sides
7. 7. Divide both sides by 6</li></li></ul><li>Fixed and Variable Cost<br />Cab fare in Metropolis is \$2.50, plus \$2.00 per mile, with a surcharge of \$1.00 for each passenger in addition to the first one. Three friends wish to take a cab to the theater. Each is willing to spend \$7.00 toward the cab ride. They figure that \$3.00 out of the total will be allocated for the tip. How far can they go on the money they have budgeted for the ride?<br />FIND: the number of miles they can travel, we’ll represent that as x.<br />FACTS: 3 people have \$7 each, for a total of \$21. However \$3 is allocated for tip, so that leaves \$18 for actual cab fare. Cost of cab is \$2.50 + \$2 per mile + \$2 for the extra passengers. So fixed cost will be \$4.50 and variable cost will be 2x (\$2 for each mile)<br />FORMULA: 2x + 4.5 = 18<br />SOLVE: 2x + 4.5 = 18<br /> 2x + 4.5 – 4.5 = 18 – 4.5<br /> 2x = 13.5<br /> x = 6.75<br />ANSWER: We used x to represent number of miles, so that means the three friends can go 6.75 miles. Does this make sense? It seems reasonable. Check your answer by substituting 6.75 into the original equation: (2)(6.75) + 4.5 = 18 <br /> 13.5 + 4.5 = 18<br /> 18 = 18 <br /><ul><li>Subtract 4.5 from each side
8. 8. Divide both sides by 2</li></li></ul><li>Percent of Increase or Decrease<br />John bought a sweater at 40% off. He paid \$29.97. What was the original price?<br />FIND: original price, x<br />FACTS: discount is 40% of the original price or .40x; sale price = original price – discount <br />FORMULA: x - .40x = 29.97<br />SOLVE: x - .40x = 29.97<br />.6 x = 29.97<br /> x = 49.95<br />ANSWER: Does this answer make sense? The item was on sale, so the original price should be more. This answer looks reasonable. Substitute it into the original equation. Does it check?<br /><ul><li>Combine like terms
9. 9. Divide both sides by .6</li>