1 Use the problem-solving procedure.
2 Solve for a variable in an equation or formula.
1 Use the Problem-Solving Procedure
One of the main reasons for studying mathematics is that we can use it to solve everyday problems. To solve most real-life application problems mathematically, we need to be able to express the problem in mathematical symbols using expressions or equations, and when we do this, we are creating a mathematical model of the situation.
In this section, we present a problem-solving procedure and discuss formulas. A for- mula is an equation that is a mathematical model of a real-life situation. Throughout the book we will be problem solving. When we do so, we will determine an equation or for- mula that represents or models the real-life situation.
You can approach any problem using the general five-step problem-solving proce- dure developed by George Pólya and presented in his book How to Solve It.
Guidelines for Problem Solving 1. Understand the problem.
• Read the problem carefully at least twice. In the first reading, get a general overview of the problem. In the second reading, determine (a) exactly what you are being asked to determine and (b) what information the problem provides.
• If possible, make a sketch to illustrate the problem. Label the information given.
• List the information in a table if it will help in solving the problem.
(Continued)
The following examples show how to apply the guidelines for problem solving.
When necessary, we will provide the steps in the examples to illustrate the five-step pro- cess. As was stated in step 2 of the problem-solving process—translate the problem into mathematical language—we will sometimes need to select and use a formula. We will show how to do that in this section.
The first formula we will discuss is the simple interest formula:
interest = principal#rate#time or i = prt where i = the simple interest
p = the principal 1the original amount of money borrowed2 r = interest rate written in decimal form
t = time
Note that the rate and the time must both use the same time measure. We will usually use years.
Next, we will work an example that uses the simple interest formula.
2. Translate the problem into mathematical language.
• This will generally involve expressing the problem algebraically.
• Sometimes this involves selecting a particular formula to use, whereas other times it is a matter of generating your own equation. It may be necessary to check other sources for the appropriate formula to use.
3. Carry out the mathematical calculations necessary to solve the problem.
4. Check the answer found in step 3.
• Ask yourself: “Does the answer make sense?” “Is the answer reasonable?” If the answer is not reasonable, recheck your method for solving the problem and your calculations.
• Check the solution in the original problem if possible.
5. Answer the question. Make sure you have answered the question asked. State the answer clearly in a sentence.
EXAMPLE 1 A Personal Loan Gwen makes a $2500, 3.5% simple interest loan to her friend, Eric, for a period of 4 years.
a) At the end of 4 years, what is the amount of simple interest that Eric will pay Gwen?
b) When Eric settles his loan, how much money, in total, must he pay Gwen?
Solution a) Understand When a simple interest loan is made, the borrower must pay both the simple interest and the principal when the loan is settled. Here the princi- pal is $2500, the simple interest rate is 3.5%, or 0.035, and the time is 4 years.
Translate We will use the simple interest formula, i = prt. In this problem, p = 2500, i = 0.035, and t = 4.
Carry Out i = prt
i = 250010.0352142 i = 350
Check The answer appears reasonable in that Eric will pay $350 for the use of $2500 for 4 years.
Answer The simple interest owed is $350.
b) Eric must repay the principal, $2500, plus the simple interest determined in part a),
$350. Thus, Eric must pay Gwen, $2500 + $350, or $2850.
Now Try Exercise 67 Understanding Algebra
When converting a number written as a percent to a decimal number, remember that percent means per 100 or divided by 100. Thus,
3.5% = 3.5
100 =0.035.
When you deposit money into a bank account you are likely to be paid compound interest. This means that you get interest on your investment for one period of time. Then in the next period you get interest paid on your investment, plus you get interest paid on the interest that was paid in the first period. This process continues for each period. The compound interest formula is:
A = p a1 + r nbnt
where A = the accumulated amount, or the balance, in the account p = the principal, or the initial investment
r = the interest rate written in decimal form
n = the number of times per year the interest is compounded t = the time of the investment, measured in years
Our next example will involve using the compound interest formula.
EXAMPLE 2 Certificate of Deposit Pola Sommers received a holiday bonus of
$1350 and invests the money in a certificate of deposit (CD) at a 3.6% annual interest rate compounded monthly for 18 months.
a) How much money will the CD be worth in 18 months?
b) How much interest will she earn during the 18 months?
Solution a) Understand and Translate We will use the compound interest formula, A = pa1 + r
nbnt. In this problem, we have p = $1350, r = 3.6% or 0.036, n = 12 (since there are 12 months in a year), and t = 1.5 a18 months = 18
12 = 1.5 yearsb. Substitute these values into the formula and evaluate.
A = p a1 + r nbnt Carry Out = 1350a1 + 0.036
12 b1211.52
= 135011 + 0.003218
= 135011.003218
≈ 135011.055399282 From a calculator
≈ 1424.79 Rounded to the nearest cent
Check The answer $1424.79 is reasonable, since it is more than Pola originally invested.
Answer Pola’s CD will be worth $1424.79 at the end of 18 months.
b) Understand The interest will be the difference between the original amount in- vested and the value of the certificate of deposit at the end of 18 months.
Translate interest = ¢ value of the certificate
of deposit after 18 months≤ - ¢amount originally invested ≤ Carry Out = 1424.79 - 1350 = 74.79
Check The amount of interest is reasonable and the arithmetic is easily checked.
Answer The interest gained in the 18-month period will be $74.79.
Now Try Exercise 81 Our next example involves a formula containing variables with subscripts. Subscripts are numbers (or other variables) placed below and to the right of variables. In Example 3, we use the variable v0. In the variable v0, the 0 is a subscript. Subscripts are read using the word “sub.” For example, v0 is read “v sub zero.”
2 Solve for a Variable in an Equation or Formula
There are many occasions when you might be given an equation or formula that is solved for one variable but you want to solve it for a different variable. Since formulas are equa- tions, the same procedure we use to solve for a variable in an equation will be used to solve for a variable in a formula.
When you are given an equation (or formula) that is solved for one variable and you want to solve it for a different variable, treat each variable in the equation, except the one you are solving for, as if it were a constant. Then isolate the variable you are solving for using the procedures similar to those previously used to solve equations.
EXAMPLE 3 Height of a Tennis Ball When an object is projected directly upward, the height, h, of the object above the ground at time, t, can be determined using the following formula:
h = -16t2 + v0t + h0
where h is the height of the object in feet, t is the time in seconds, v0 is the initial or start- ing velocity of the object, and h0 is the initial or starting height of the object. A tennis ball is projected directly upward with an initial velocity of 40 feet per second from an initial height of 3 feet.
a) Determine the height of the tennis ball after 1 second.
b) Determine the height of the tennis ball after 2.5 seconds.
Solution Understand and Translate In this problem, the height of the tennis ball can be determined using the given formula. Our first step is to identify the numbers to substitute into the formula for v0 and h0. The initial velocity of the tennis ball is 40 feet per second, therefore v0 = 40. The initial height of the tennis ball is 3 feet, therefore h0 = 3. We rewrite the formula substituting in these values:
h = -16t2 + v0t + h0
h = -16t2 + 40t + 3
To determine the height of the tennis ball in parts a) and b) we will substitute the values of t into this equation.
Carry Out
a) The height of the tennis ball after 1 second can be determined by substituting t = 1 into the equation.
h = -16t2 + 40t + 3 h = -161122 + 40112 + 3
= -16#1 + 40 + 3 = -16 + 40 + 3 = 27 Thus, the height of the tennis ball after 1 second is 27 feet.
b) The height of the tennis ball after 2.5 seconds can be determined by substituting t = 2.5 into the equation.
h = -16t2 + 40t + 3
h = -1612.522 + 4012.52 + 3
= -1616.252 + 100 + 3
= -100 + 100 + 3 = 3
Thus, the height of the tennis ball after 2.5 seconds is 3 feet.
Now Try Exercise 85
EXAMPLE 4 Solve the equation 5x - 8y = 32 for y.
Solution We will solve for the variable y by isolating the term containing the y on the left side of the equation.
5x - 8y = 32
5x - 5x - 8y = -5x + 32 Subtract 5x from both sides.
-8y = -5x + 32 -8y
-8 = -5x + 32
-8 Divide both sides by -8.
y = -5x + 32 -8
y = -11-5x + 322 -11-82
Multiply the numerator and the denominator by -1.
y = 5x - 32
8 or y = 5
8 x - 4
Now Try Exercise 27
EXAMPLE 5 Solve the equation y + 1 = 2
51x - 42 for y.
Solution We begin by multiplying both sides of the equation by the least common denominator, 5.
y + 1 = 2
51x - 42 51y + 12 = 5c2
51x - 42 d Multiplied both sides by LCD, 5.
51y + 12 = 21x - 42
5y + 5 = 2x - 8 Used distributive property.
5y + 5 - 5 = 2x - 8 - 5 Subtract 5 from both sides.
5y = 2x - 13 5y
5 = 2x - 13
5 Divide both sides by 5.
y = 2x - 13
5 or y = 2
5 x - 13 5
Now Try Exercise 35 Next, we will solve for a variable in a formula. Remember: Our goal is to isolate the variable for which we are solving.
EXAMPLE 6 Rectangle Perimeter The formula for the perimeter of a rectangle is P = 2l + 2w, where l is the length and w is the width of the rectangle (see Fig. 2.2).
Solve this formula for the width, w.
Solution Since we are solving for w, we must isolate the w on one side of the equation.
P = 2l + 2w
P - 2l = 2l - 2l + 2w Subtract 2l from both sides.
P - 2l = 2w P - 2l
2 = 2w
2 Divide both sides by 2.
P - 2l
2 = w
Rectangle
l w
FIGURE 2.2
Thus, w = P - 2l
2 or w = P 2 - 2l
2 = P 2 - l.
Now Try Exercise 49
EXAMPLE 7 Trapezoid Area A formula used to determine the area of a trapezoid is A = 1
2 h1b1 + b22, where h is the height and b1 and b2 are the lengths of the bases of the trapezoid (see Fig. 2.3). Solve this formula for b2.
Solution We begin by multiplying both sides of the equation by the LCD, 2, to clear fractions.
A = 1
2 h1b1 + b22 2#A = 2 c1
2 h1b1 + b22 d Multiply both sides by 2.
2A = h1b1 + b22 2A
h = h1b1 + b22
h Divide both sides by h. 2A
h = b1 + b2 2A
h - b1 = b1 - b1 + b2 Subtract b1 from both sides.
2A
h - b1 = b2
Now Try Exercise 57 Trapezoid
b1
b2 h
FIGURE 2.3
EXAMPLE 8 Charles’s Law A formula used in chemistry that relates the volume and temperature of a gas is V1
T1 = V2
T2, where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature. This formula is known as Charles’s Law: Solve this formula for T2.
Solution We begin by multiplying both sides of the equation by the LCD of the frac- tions, T1T2.
T1T2#aV1
T1b = T1T2#aV2
T2b Multiply both sides by T1T2. T2V1 = T1V2
T2V1 V1 = T1V2
V1 Divide both sides by V1
T2 = T1V2 V1
Now Try Exercise 55