§ 1.5 Problem Solving and Using Formulas

Blitzer, Algebra for College Students, 6e – Slide #2 Section 1.5 Solving Word Problems Strategy for Solving Word Problems STEP 1 Read the problem carefully. Attempt to state the problem in your own words and state what the problem is looking for. Let x (or any variable) represent one of the quantities in the problem. STEP 2 If necessary, write expressions for any other unknown quantities in the problem in terms of x. STEP 3 Write an equation in terms of x that describes the verbal conditions of the problem. (translate from English to “Math”) 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, Algebra for College Students, 6e – Slide #3 Section 1.5 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 “terror”. 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 that problem that was written in words.

Blitzer, Algebra for College Students, 6e – Slide #4 Section 1.5 Solving Word ProblemsEXAMPLE SOLUTION 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? 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, Algebra for College Students, 6e – Slide #5 Section 1.5 Solving Word ProblemsCONTINUED 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, Algebra for College Students, 6e – Slide #6 Section 1.5 Solving Word ProblemsCONTINUED STEP 4: Solve the equation and answer the question. 4x + 2x = 300Multiply 6x = 300Add like terms x = 50 Divide both sides by 6 2(2x) + 2(x) = 300 Therefore the width of the soccer field is 50 yards. Also, I therefore know that 2x = 2(50) = 100. Therefore, the length of the soccer field is 100 yards.

Blitzer, Algebra for College Students, 6e – Slide #7 Section 1.5 Solving a Formula for a Variable We know that solving an equation is the process of finding the number or number that makes the equation a true statement. Formulas contain two or more letters, representing two or more variables. The formula for the perimeter P of a rectangle is 2l +2w = P where l is the length and w is the width of the rectangle. We say that the formula is solved for P, since P is alone on one side and the other side does not contain a P.

Blitzer, Algebra for College Students, 6e – Slide #8 Section 1.5 Solving a Formula for a Variable Solving a formula for a variable means using the addition and multiplication properties of equality to rewrite the formula so that the variable is isolated on one side of the equation. To solve a formula for one of its variables, treat that variable as if it were the only variable in the equation. Think of the other variables as if they were just numbers. Use the addition property of equality to isolate all terms with the specified variable on one side. Then use the multiplication property of equality to get the specified variable alone. The next example shows how to do this.

Blitzer, Algebra for College Students, 6e – Slide #9 Section 1.5 Solving a Formula for a VariableEXAMPLE SOLUTION Solve the formulafor h.

Blitzer, Algebra for College Students, 6e – Slide #10 Section 1.5 Solving Word ProblemsEXAMPLE SOLUTION A coupon book for a bridge costs $30 per month. The toll for the bridge is normally $5, but it is reduced to $3.50 for people who have purchased the coupon book. Determine the number of times in a month the bridge must be crossed so that the total monthly cost without the coupon book is the same as the total monthly cost with the coupon book. STEP 1: Let x represent one of the quantities. x = number of times someone crosses the bridge in a month

Blitzer, Algebra for College Students, 6e – Slide #11 Section 1.5 Solving Word ProblemsCONTINUED STEP 3: Write an equation in x that describes the conditions. The amount of money spent crossing the bridge without the coupon book is 5x. STEP 2: Represent other unknown quantities in terms of x. There are no other unknown quantities, so we can skip this step. The amount of money spent crossing the bridge with the coupon book is x (30 is not multiplied by x since it is a fixed cost).

Blitzer, Algebra for College Students, 6e – Slide #12 Section 1.5 Solving Word ProblemsCONTINUED STEP 4: Solve the equation and answer the question. Subtract 3.5x from both sides Divide both sides by 1.5 Therefore, after using the coupon book 20 times, it becomes profitable. Since we want to know when the it begins to be profitable to use the coupon book, we want to know when the two methods of paying the bridge toll are equal. Therefore the equation to be used is 5x = x. 5x = x 1.5x = 30 x = 20

Blitzer, Algebra for College Students, 6e – Slide #13 Section 1.5 Solving Word ProblemsCONTINUED Because the only unknown quantity is the variable for which we are solving, it is not necessary to check our answer. However, it is always a very good idea to verify our calculations. STEP 5: Check the proposed solution in the original wording of the problem. Verify the calculations on your own right now and then say the following ten times: “I can solve word problems!”