§ 2.5

1. §2.5 An Introduction to Problem Solving

2. Getting Started – A Strategy and Some Terminology The hardest thing about word problems is writing an equation that models the problem’s conditions. The following tables summarize many of the algebraic expressions that you have used in previous sections for modeling the conditions. We will use “x” for the variable, but you can use another letter if you prefer. Blitzer, Introductory Algebra, 5e – Slide #2 Section 2.5

3. Strategy for Solving Word Problems Again… Step 1 Read the problem carefully. Attempt to state what the problem is looking for. Let x (or any variable) represent one of the unknown quantities. Step 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x. Blitzer, Introductory Algebra, 5e – Slide #3 Section 2.5

4. Strategy for Solving Word Problems Step 3 Write an equation in x that describes the conditions of the problem. Step 4 Solve the equation and answer the problem’s question. Step 5 Check the solution in the original wording of the problem, not in the equation obtained from the words. Blitzer, Introductory Algebra, 5e – Slide #4 Section 2.5

5. Algebraic Translations of English Phrases Addition Blitzer, Introductory Algebra, 5e – Slide #5 Section 2.5

6. Algebraic Translations of English Phrases Subtraction Pay careful attention to which of the terms should have the minus sign in these examples. It’s all in the wording! Blitzer, Introductory Algebra, 5e – Slide #6 Section 2.5

7. Algebraic Translations of English Phrases Division Blitzer, Introductory Algebra, 5e – Slide #7 Section 2.5

8. Practice The product of 5 and a number is 135. Find the number. EXAMPLE • Let x represent one of the quantities. x = the number. • Represent other quantities in terms of x. There are no other unknown quantities. • Write an equation in x that describes the conditions. 5x = 135 • Solve the equation and answer the question. x = 27 Divide both sides by 5. The number is 27. • Check the solution. The product of 5 and 27 is 135. Blitzer, Introductory Algebra, 5e – Slide #8 Section 2.5

9. Algebraic Translations of English Phrases Consecutive Integers Blitzer, Introductory Algebra, 5e – Slide #9 Section 2.5

10. Practice EXAMPLEThe sum of three consecutive odd integers is 45. Find the integers. • Let x represent one of the quantities. x = the number. • Represent other quantities in terms of x. x + 2 = the second number, x + 4 = the third number • Write an equation in x that describes the conditions. x + x + 2 + x + 4 = 45 • Solve the equation and answer the question. 3x + 6 = 45 Combine like terms. 3x = 39 Subtract 6 from both sides. x =13, x + 2 = 15, x + 4 =17 The numbers are 13, 15 and 17. • Check the solution. The sum of 13, 15 and 17 is 45. Blitzer, Introductory Algebra, 5e – Slide #10 Section 2.5

11. Practice EXAMPLE :A field is six times as long as it is wide. If the perimeter is 420 feet, find the dimensions of the field. • Let w represent one of the quantities. w = the width of the field • Represent other quantities in terms of w. 6w = length • Write an equation in w that describes the conditions. Recall: 2w + 2l = P 2w + 2(6w) = 420 • Solve the equation and answer the question. 2w+12w = 420 Multiply. 14w = 420 Combine like terms. w = 30 and 6w = 180 Solve. Answer: The width is 30 feet and the length is 180 feet. Check the solution. The perimeter is 2(30) + 2(180) = 60 +360 = 420 feet. Blitzer, Introductory Algebra, 5e – Slide #11 Section 2.5

12. Solving Word Problems – Remember your strategy One more time… Blitzer, Introductory Algebra, 5e – Slide #12 Section 2.5

13. Solving Word Problems Study Tip When solving word problems, particularly problems involving geometric figures, drawing a picture of the situation is often helpful. Label x on your drawing and where appropriate, label other parts of the drawing in terms of x. For some people, just hearing “word problem” is about as unsettling as hearing words like “cancer” or “terrorist”. But remember – real world problems come in words and if you are to use your algebra – you must learn to make word problems your friends. You can do it! Read the problem slowly, draw a picture, think of what you are looking for and call it “x”, and you will be well on your way to solving the problem that was written in words. Blitzer, Introductory Algebra, 5e – Slide #13 Section 2.5

14. Solving Word Problems EXAMPLE A rectangular soccer field is twice as long as it is wide. If the perimeter of the soccer field is 300 yards, what are its dimensions? SOLUTION STEP 1: Let x represent one of the quantities. x = the width of the soccer field. STEP 2: Represent other unknown quantities in terms of x. Since the field is twice as long as it is wide, then 2x = the length of the soccer field. Blitzer, Introductory Algebra, 5e – Slide #14 Section 2.5

15. Solving Word Problems CONTINUED STEP 3: Write an equation in x that describes the conditions. The soccer field is in the shape of a rectangle and therefore has a perimeter equal to twice the length plus twice the width. This can be expressed as follows: 2(2x) + 2(x) = 300 Blitzer, Introductory Algebra, 5e – Slide #15 Section 2.5

16. Solving Word Problems CONTINUED STEP 4: Solve the equation and answer the question. 2(2x) + 2(x) = 300 4x + 2x = 300 Multiply 6x = 300 Add like terms Divide both sides by 6 x = 50 Therefore the width of the soccer field is 50 yards. The length of the field is twice the width, and 2x = 2(50) = 100. Therefore, the length of the soccer field is 100 yards. Blitzer, Introductory Algebra, 5e – Slide #16 Section 2.5

17. Solving Word Problems CONTINUED STEP 5: Check the proposed solution in the original wording of the problem. The problem states that the perimeter of the soccer field is 300 yards. Let’s use this information to verify our answer. The formula for the perimeter of a rectangle is repeated as follows: 2(2x) + 2(x) = 300 2(2(50)) + 2(50) = 300 Replace x with 50 200 + 100 = 300 Multiply 300 = 300 Add So, the dimensions of the soccer field are 50 yards by 100 yards. Blitzer, Introductory Algebra, 5e – Slide #17 Section 2.5