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Section 1.4 Solving Equations

Subsection What is an Equation?

An equation is a statement that two expressions are equal. It may involve one or more variables.

Example 1.24.

Liz makes $6 an hour as a tutor in the Math Lab. Her weekly earnings, \(w\text{,}\) depend on the number of hours she works, \(h\text{.}\) The equation relating these two variables is

\begin{equation*} w=6h \end{equation*}

If we know the value of \(h\text{,}\) we can evaluate the expression \(6h\) to find Liz's earnings. For example, if \(h=\alert{7}\text{,}\) then

\begin{equation*} w=6(\alert{7})=42 \end{equation*}

Liz makes $42 for 7 hours of work.

In Example 1.24 we used evaluation to find the value of \(w\) when we knew the value of \(h\text{.}\) What if we know the value of \(w\) and want to find \(h\text{?}\) Suppose Liz earned $54 last week. How many hours did she work? To answer this question we substitute \(\alert{54}\) for \(w\text{,}\) so that the equation becomes

\begin{equation*} \alert{54} = 6h \end{equation*}

Now we have an equation in just one variable. We would like to find the value of \(h\) that makes this equation true.

A value of the variable that makes an equation true is called a solution of the equation, and the process of finding this value is called solving the equation.

Subsection Trial and Error

Some equations are easy enough to solve by trial and error. You can probably see that the solution of the equation \(54=6h\) is 9. We can use a table to help us solve a harder equation by trial and error.

Exercise 1.25.

One of Aunt Esther's Chocolate Dream cookies contains 42 calories, so \(d\) cookies contain \(c\) calories, and \(c=42d\text{.}\) If Albert consumed 546 calories, how many cookies did he eat? Use trial and error to help you solve the equation \(546=42d\text{.}\)

\(d\) \(5\) \(6\) \(7\) \(8\) \(9\) \(10\) \(11\) \(12\) \(13\) \(14\) \(15\)
\(c\) \(\hphantom{00}\) \(\hphantom{00}\) \(\hphantom{00}\) \(\hphantom{00}\) \(\hphantom{00}\) \(\hphantom{00}\) \(\hphantom{00}\) \(\hphantom{00}\) \(\hphantom{00}\) \(\hphantom{00}\) \(\hphantom{00}\)

Look Ahead.

Trial and error can be time-consuming, especially for complicated equations! In this section we investigate some algebraic methods for solving equations.

Reading Questions Reading Questions

1.

What is an equation?

2.

What is a solution of an equation?

Subsection Inverse Operations

Instead of relying on trial and error to solve equations, we will use inverse operations. Try the Exercise below.

Exercise 1.26.
\begin{equation*} \begin{aligned} \amp \blert{\text{Choose any number for}~x:} \amp x \amp = ~\rule{1cm}{0.15mm}\\ \amp \blert{\text{Multiply your number by 5}:} \amp 5x \amp = ~\rule{1cm}{0.15mm}\\ \amp \blert{\text{Divide the result by 5}:} \amp \dfrac{5x}{5} \amp = ~\rule{1cm}{0.15mm} \end{aligned} \end{equation*}

Did you end up with your original number? Try multiplying and dividing your number by another number besides 5. Do you still end up with your original number?

Multiplication and division are opposite or inverse operations, because each operation undoes the effects of the other.

Exercise 1.27.
\begin{equation*} \begin{aligned} \amp \blert{\text{Choose any number for}~x:} \amp x \amp = ~\rule{1cm}{0.15mm}\\ \amp \blert{\text{Add 4 to your number}:} \amp x+4 \amp = ~\rule{1cm}{0.15mm}\\ \amp \blert{\text{Subtract 4 from the result}:} \amp x+4-4 \amp = ~\rule{1cm}{0.15mm} \end{aligned} \end{equation*}

Did you end up with your original number? Try adding and subtracting another number besides 4. Do you still end up with your original number?

Addition and subtraction are opposite or inverse operations, because each operation undoes the effects of the other.

We see that the opposite operation for multiplication is division, and the opposite operation for addition is subtraction. Now we can solve the equation in Example 1.24.

Example 1.28.

Solve the equation \(54=6h\text{.}\)

Solution

We will transform this equation into a new one of the form \(h=k\) (or \(k=h\)), where the number \(k\) is the solution.

To do this, we must isolate the variable \(h\) on one side of the equals sign. Because \(h\) is multiplied by 6 in this equation, we divide both sides by 6, like this:

\begin{equation*} \begin{aligned} \dfrac{54}{\blert{6}} \amp = \dfrac{6h}{\blert{6}} \amp \amp \blert{\text{Divide both sides by 6.}}\\ 9 \amp = h \end{aligned} \end{equation*}

The solution is 9. Liz worked for 9 hours.

Look Closer.

Note that the expression \(\dfrac{6h}{6}\) says "multiply \(h\) by 6, then divide the result by 6," so we end up with \(h\) again.

In Example 1.28, we transformed the equation into a simpler one (namely, \(h=9\)) which tells us the solution. We can use the following operations to transform an equation:

To solve an equation algebraically.
  1. We can add or subtract the same number from both sides.
  2. We can multiply or divide both sides by the same number, as long as that number is not zero.
Look Closer.

Our method for solving equations works because of the properties of equality, one for each of the four arithmetic operations. They are listed below.

Addition and Subtraction Properties of Equality.

If the same quantity is added to or subtracted from both sides of an equation, the solution is unchanged. In symbols,

\begin{equation*} \blert{\text{If}~~a=b,~~ \text{then}~~a+c=b+c~~\text{and}~~a-c=b-c} \end{equation*}
Multiplication and Division Properties of Equality.

If both sides of an equation are multiplied or divided by the same nonzero quantity, the solution is unchanged. In symbols,

\begin{equation*} \blert{\text{If}~~a=b,~~ \text{then}~~ac=bc~~\text{and}~~\dfrac{a}{c}=\dfrac{b}{c},~~c\not=0} \end{equation*}

Applying any one of these properties to an equation produces an equivalent equation: one with the same solutions as the original equation.

Subsection Solving Equations

Now we can solve equations algebraically. We'll start with the simple Example 1.17 from Section 1.3 about Delbert's and Francine's ages.

Example 1.29.

Francine is exactly four years older than Delbert, so an equation relating their ages is \(F=D+4\text{.}\) How old is Delbert when Francine is 19?

Solution

We must solve the equation

\begin{equation*} 19=D+4 \end{equation*}

We see that 4 has been added to the variable, \(D\text{.}\) To isolate \(D\) on one side of the equals sign, we subtract 4 from both sides of the equation, like this:

\begin{equation*} \begin{aligned} 19\, \amp = D+4 \\ \underline{\blert{-4}} \amp = \underline{~~~\,\blert{-4}} \amp \amp \blert{\text{Note that} ~D+4-4=D}\\ 15 \amp = D \end{aligned} \end{equation*}

The solution to this equation is 15. You can check that substituting \(\alert{15}\) for \(D\) does make the equation true.

\begin{equation*} \begin{aligned} \blert{\text{Check:}}\amp \amp 19 \amp = D+4 \amp \amp \blert{\text{Substitute}~ 15~ \text{for}~D.}\\ \amp \amp 19 \amp =\alert{15} +4 \amp \amp \blert{\text{True: the solution checks.}} \end{aligned} \end{equation*}
Look Ahead.

That last example is easy enough to solve without algebra, but the method will help us with harder problems. Here is a general strategy for solving such equations.

To solve an equation algebraically.
  1. Ask yourself what operation has been performed on the variable.
  2. Perform the opposite operation on both sides of the equation in order to isolate the variable.

Reading Questions Reading Questions

3.

What is the opposite operation for subtraction? For multiplication?

4.

What are equivalent equations?

Subsection Problem Solving with Equations

Some problems can be solved by applying one of our familiar formulas.

Example 1.30.

A saving's account at Al's Bank earns 5% annual interest. Jan's account earned $42.50 in interest last year. What was her balance at the beginning of the year?

Solution

First, we choose the appropriate formula. This is a problem about interest, so we use the interest formula:

\begin{equation*} \blert{I=Prt} \end{equation*}

Next, we list the values of the variables. (Which variable is unknown?)

\begin{equation*} \begin{aligned} I \amp = \alert{42.50} \amp r \amp = \alert{0.05}\\ P \amp = \text{unknown} \amp t \amp = \alert{1} \end{aligned} \end{equation*}

We substitute the values into the formula:

\begin{equation*} \alert{42.50}=P(\alert{0.05})(\alert{1}) \end{equation*}

Then we solve the equation:

\begin{equation*} \begin{aligned} \dfrac{42.50}{0.05} \amp =\dfrac{P(0.05)}{0.05} \amp \amp \blert{\text{Divide both sides by}~0.05.}\\ 850 \amp = P \end{aligned} \end{equation*}

Finally, we answer the question in a sentence:

\begin{equation*} \blert{\text{Jan's account balance was }\$850 \text{ at the beginning of the year.}} \end{equation*}

Many practical problems can be solved by first writing an equation that describes or models the problem, and then solving the equation. We will use three steps to solve applied problems.

Steps for solving an applied problem:.
  1. Identify the unknown quantity and choose a variable to represent it.
  2. Find some quantity that can be described in two different ways, and write an equation using the variable to model the problem situation.
  3. Solve the equation and answer the question in the problem.
Example 1.31.

Jerry needs an additional $35 for airfare to New York. The ticket to New York costs $293. How much money does Jerry have?

Solution

We follow the three steps.

  1. The amount of money Jerry has is unknown.
    \begin{equation*} \blert{\text{Amount of money Jerry has:}~~m} \end{equation*}
  2. The airfare to New York can be described in two different ways:

    \begin{equation*} \begin{alignedat}{5} \amp \quad m\amp +\amp\quad 35 \amp =\quad \amp ~~ 293{}\\ \amp \,\text{amount }\amp\amp~\text{amount} \amp\amp\text{airfare} \\ \amp\text{Jerry has }\amp\amp\text{he needs} \end{alignedat} \end{equation*}
  3. We solve the equation and answer the question.

    \begin{equation*} \begin{alignedat}{9} m \amp{}+ 35\amp \amp{}= ~293 \amp\amp \qquad \blert{\text{Subtract 35 from both sides.}}\\ \amp\underline{{}-{}\blert{35}} \amp \amp\quad \underline{{}-{}\blert{35}}\\ \amp \quad\, m \amp \amp {}= ~258 \end{alignedat} \end{equation*}

    Jerry has $258.

Caution 1.32.

In Example 1.31, each side of the equation represents the airfare to New York, not the amount of money Jerry has, which is what we want to find. This is typical of applied problems: the equation does not usually represent the unknown quantity itself, but some related quantity.

Look Ahead.

Although you may be able to solve the problems in this lesson with arithmetic, the important thing is to learn to write the algebraic equation, or model, for the problem. This skill will help you to solve problems that are too difficult to solve with arithmetic alone.

Reading Questions Reading Questions

5.

What is the first step in solving an applied problem?

6.

True or false: One side of the equation that models a problem will be the unknown variable.

Subsection Skills Warm-Up

Exercises Exercises

In Exercises 1–6, choose the correct algebraic expression.

\begin{align*} m+15\amp\amp m-15\amp\amp 15-m \end{align*}
1.

Carol weighs 15 pounds less than Garth. If Garth weighs \(m\) pounds, how much does Carol weigh?

2.

Amber and Beryl together planted 15 trees. If Amber planted \(m\) trees, how many trees did Beryl plant?

3.

Fred earned $15 more this week than last week. If he earned \(m\) dollars last week, how much did he earn this week?

4.

Meg bicycled 15 miles farther than Kwan. If Kwan rode \(m\) miles, how far did Meg ride?

5.

There are 15 children in Amy's swim class. If there are \(m\) girls, how many are boys?

6.

The sale price of a sweater is $15 less than the regular price. If the regular price is \(m\) dollars, what is the sale price?

In Exercises 7–12, choose the correct algebraic expression.

\begin{align*} 12p\amp\amp\dfrac{12}{p}\amp\amp \dfrac{p}{12} \end{align*}
7.

Julian earns $12 an hour. If he works for \(p\) hours, how much will he make?

8.

Farmer Brown collected \(p\) eggs this morning. How many dozen is that?

9.

Melissa bought 12 colored markers. If their total cost was \(p\) dollars, how much did each marker cost?

10.

Rosalind is baby-sitting for \(p\) children. If she brings 12 puzzles, how many will each child get?

11.

Hector has to read 12 chapters in his history text. If he has \(p\) days to complete the assignment, how many chapters should he read per day?

12.

Roma swims 12 laps per day. After \(p\) days, how many laps has she swum?

Solutions Answers to Skills Warm-Up

Exercises Exercises

Exercises Homework 1.4

For Problems 1–2, fill in the table for each equation. Explain how you found the unknown values.

1.

\(q=9+t\)

\(t\) \(2\) \(4\) \(\hphantom{00}\) \(\hphantom{00}\)
\(q\) \(\hphantom{00}\) \(\hphantom{00}\) \(15\) \(18\)
2.

\(p=5n\)

\(n\) \(0\) \(2\) \(\hphantom{00}\) \(\hphantom{00}\)
\(p\) \(\hphantom{00}\) \(\hphantom{00}\) \(20\) \(35\)
3.

Decide whether the given value for the variable is a solution of the equation.

  1. \(x-4=6;~~x=10\)
  2. \(4y=28;~~y=24\)
  3. \(\dfrac{0}{z}=0;~~z=19\)

For Problems 4–12, solve and check your solution.

4.
\(x-3=11\)
5.
\(10.6=7.8+y\)
6.
\(3y=108\)
7.
\(42=3.5b\)
8.
\(2.6=\dfrac{a}{1.5}\)
9.
\(x-4=0\)
10.
\(34x=212\)
11.
\(6z=20\)
12.
\(9=k+9\)

For Problems 13–16,

  1. Choose the appropriate formula to write an equation.
  2. Solve the equation and answer the question.
13.

Clive loaned his brother some money to buy a new truck, and his brother agreed to repay the loan in 1 year with 3% interest. Clive earned $75 interest on the loan. How much did Clive loan his brother?

14.

Andy's average homework score on eight assignments was 38.25. How many homework points did Andy earn altogether?

15.

How long will it take a cyclist traveling at 13 miles an hour to cover 234 miles?

16.

A roll of carpet contains 400 square feet of carpet. If the roll is 16 feet wide, how long is the piece of carpet?

For Problems 17–22, choose the appropriate equation.

\begin{align*} x+7\amp = 26 \amp 7x \amp = 26 \amp \dfrac{x}{7} \amp = 26\\ x-7\amp = 26 \amp \dfrac{x}{26} \amp = 7 \amp \dfrac{26}{x} \amp = 7 \end{align*}
17.

Sarah drove 7 miles farther to her high school reunion than Jenni drove. If Sarah drove 26 miles, how far did Jenni drive?

18.

Lurline and Rozik live 26 miles apart. They meet at a theme park between their homes. If Lurline drove 7 miles to the park, how far did Rozik drive?

19.

Doris is training for a triathlon. This week she averaged 26 miles per day on her bicycle. If she rode every day, what was her total mileage?

20.

Glynnis jogged the same route every day this week for a total of 26 miles. How long is her route?

21.

Astrid divided her supply of colored pencils among the 26 children in her class, and each child got 7 pencils. How many pencils does she have?

22.

Ariel lost 7 of the beads on her necklace, and now there are 26. How many were there originally?

Follow the steps to solve Problems 23–26.

23.

Lupe spent $24 at the Craft Fair. She now has $39 left. How much did she have before the Craft Fair?

  1. What are we asked to find? Choose a variable to represent it.
  2. Find two ways to express the amount of money Lupe had after the Craft Fair, and write an equation.
  3. Solve the equation and answer the question in the problem.
24.

Danny weighs 32 pounds more than Brenda. If Danny weighs 157 pounds, how much does Brenda weigh?

  1. What are we asked to find? Choose a variable to represent it.
  2. Find two ways to express Danny's weight, and write an equation.
  3. Solve the equation and answer the question in the problem.
25.

Miranda worked 20 hours this week and made $136. What is Miranda's hourly wage?

  1. What are we asked to find? Choose a variable to represent it.
  2. Find two ways to express Miranda's total earnings, and write an equation.
  3. Solve the equation and answer the question in the problem.
26.

Struggling Students Gardening Service splits their profit equally among their eight members. If each member made $64 last week, what was the total profit?

  1. What are we asked to find? Choose a variable to represent it.
  2. Find two ways to express each member's share, and write an equation.
  3. Solve the equation and answer the question in the problem.

Write algebraic equations to solve Problems 27–29. Follow the steps in the Lesson.

27.

Martha paid $26 less for a suit at a discount store than her mother paid at a boutique for the same suit. If Martha paid $89 for the suit, how much did her mother pay?

28.

Emily spends 40% of her monthly income on rent. If her rent is $360 a month, how much does Emily make?

29.

After she wrote a check for $2378, Avril's bank account shows a balance of $1978. How much money was in Avril's account before the check cleared?

In Problems 30–32, we compare evaluating an expression and solving an equation.

30.

A used car costs $3400 less than the new version of the same model.

  1. Choose two variables and write an equation for the cost of the used car in terms of the cost of the new car.
  2. If the new car costs $14,500, how much does the used car cost?
  3. If the used car costs $9200, how much does the new car cost?
31.

The Dodgers won 60% of their games last season.

  1. Choose two variables and write an equation for the number of games the Dodgers won in terms of the number of games they played.
  2. If the Dodgers played 120 games, how many did they win?
  3. If the Dodgers won 96 games, how many did they play?
32.

Sunshine Industries manufactures beach umbrellas. Their profit on each umbrella is 18% of the selling price.

  1. Choose two variables and write an equation for the profit on each umbrella in terms of its selling price.
  2. If a beach umbrella sells for $60, what is the profit?
  3. If the profit on one umbrella is $7.20, what is the selling price?