# 2.3.7: Applications of Two-Step Equations

- Page ID
- 4377

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)## Solve Real-World Problems Using Rational Numbers and Simple Equations

Jill is doing a little woodwork for a school project. She had a piece of wood 4 feet long. She cut off a piece that was 2(5/8) feet long. She needs to determine how long the piece of wood was that she has left in order to see if she has enough to complete her project. How much does she have left? If she needs 5 more pieces that are each 3/10 of a foot long, does she need more wood?

In this concept, you will learn to solve real-world problems by using rational numbers and simple equations.

### Using Rational Numbers and Equations in Real World Problems

You can use fractions, decimals, and integers to solve real-world problems.

Let’s look at an example.

Candy is 5 years older than one-third Liam’s age. If Candy is 16, how old is Liam?

First, choose a variable to represent Liam’s age.

Let l represent Liam’s age.

Next, write an equation using the information in the problem.

The phrase “5 years older” translates to “+5”.

The phrase “one-third Liam’s age” translates to “(1/3)l”

Since you now know that you can describe Candy's age as "(1/3)l+5," and the problem also tells you that Candy is 16, you can write a equation. Remember that an equation is just a statement describing two equal things, using the language of mathematics.

Put the parts together to form an equation.

(1/3)l+5=16

Then, solve the equation for l. Remember that any operation you perform on one side of the equation must also be performed on the other side.

First, subtract 5 from both sides to isolate the variable term on one side of the equation.

(1/3)l+5−5=16-5

(1/3)l=11

Next, you need to eliminate a fractional coefficient, so multiply by the reciprocal. Remember the Multiplicative Inverse Property: any number multiplied by its reciprocal is equal to 1.

(3/1)(1/3)l

l=33

The answer is 33.

Liam is 33 years old.

### Examples

Example 2.3.7.1

Earlier, you were given a problem about Jill and her wood working project. She used 258feet from a 4 foot long piece of wood and needs 5 more pieces that are each 310 of a foot long. Will she have enough?

**Solution**

First, convert the mixed number into an improper fraction.

2(5/8)=((8x2)+5)/8

2(5/8)=21/8

Next, find out how much wood is left after Jill finished cutting her first piece.

x=4-(21/8)

x=(32/8)-(21/8)

x=11/8

Then, find out how much wood she needs for the 5 other pieces.

y=5x(3/10)

y=15/10

Then, compare the numbers 11/8(how much she has) and 15/10(how much she needs) to see which one is bigger. To do this, you need to get a common denominator. The common denominator for 8 and 10 is 40.

(11/8)×(5/5)=55/40

(15/10)×(4/4)=60/40

Then, compare the two fractions.

60/40>55/40

Therefore 15/10>11/8.

The answer is no, Jill does not have enough wood to cut 5 more pieces after she cuts off the first piece.

Example \(\PageIndex{1}\)

Ben bought one share of stock at 40 points. He watched the price of his stock every day for a week. On Monday, the stock moved 2.5 points up. On Tuesday, it moved 1.75 points down. On Wednesday, it moved p points. On Thursday and Friday, the stock moved 0.75 points up each day. If the price of the stock was 45 points at the end of the week, what is the value of p?

**Solution**

First, use the information in the problem to write an equation to represent the price of the stock.

40+2.5−1.75+p+2(0.75)=45

Next, use the order of operations to simplify the equation. First, multiply, then add and subtract in order from left to right.

40+2.5−1.75+p+1.5=45

42.5−1.75+p+1.5=45

40.75+p+1.5=45

42.25+p=45

Then solve for p by isolating p on the left side of the equation and simplifying the other side.

42.25+p=45

p=45-42.25

p=2.75

The answer is 2.75.

The stock moved up 2.75 points.

Example \(\PageIndex{1}\)

The product of 5 and a number is 160.

**Solution**

First, use the information in the problem to write an equation to represent the problem.

The phrase “the product of 5 and a number” translates to “5x”.

The product is equal to 160.

Therefore 5x=160

Next, solve for x by eliminating the 5 or 5/1 by multiply by the reciprocal. Remember the Multiplicative Inverse Property: any number multiplied by its reciprocal is equal to 1.

(1/5)(5x/1)=160(1/5)

x=160/5

x=32

The answer is 32.

Example \(\PageIndex{1}\)

The selling price of a certain DVD is $7 more than the price the store paid. If the selling price is $24, what did the store pay?

**Solution**

First, use the information in the problem to write an equation to represent the problem.

The phrase “$7 more” translates to “+7”.

The selling price is equal to 24.

Therefore x+7=24

Next, solve for x by subtracting 7 from both sides of the equation.

x+7−7=24-7

x=17

The answer is 17.

The store paid $17 for the DVD.

Example \(\PageIndex{1}\)

Suppose you drive 630 miles in 10.5 hours. What was your average speed?

**Solution**

First, use the information in the problem to write an equation to represent the problem. Remember that speed is equal to distance over time.

v=d/t

v=(630 miles)/(10.5 hours)

Next, solve for v.

v=(630 miles)/(10.5 hours)

v=60miles/hour

The answer is 60.

The average speed is 60 miles/hour.

### Review

Answer each question.

- The expression 2p
^{3}−4m can be used to find the sales profit of a company where p is the number of products they sell and m is the number of miles they travel. If they sold 5 products and traveled 10 miles, what was their profit? - If they sold 6 products and traveled 15 miles, what was their profit?
- If they sold 10 products and traveled 20 miles, what was their profit?
- The expression (11s/2)+7t can be used to find the group admission price, where s is the number of students and t is the number of teachers. If there are 20 students and 4 teachers, what is the group admission price?
- If there are 25 students and 4 teachers, what is the group price?
- If there are 20 students and 5 teachers, what is the group price?
- Brooke needs to save $146 for a trip. She has $35 in her savings account. She saves $15.75 each week. She also has to spend $15 to buy a present for a friend. How many weeks will Brooke need to save to have enough for her trip?
- Vinnie is 1/2 as old as Julie. Julie is 24. How old is Vinnie?
- Manuel starts with $30. He earns $8.00 per hour plus an additional bonus of $12 each day. He spends $8.00 for lunch. If he has $94 at the end of the day, for how many hours did he work?
- A formula for the perimeter of a rectangle is P=2(l+w), where P is the perimeter, l is the length, and w is the width. If the perimeter of a rectangle is 312 centimeters and the width is 67.3 centimeters, what is the length?
- If the length of a rectangle is 32 centimeters and the width is 65.5 centimeters, what is the perimeter?
- If the length of a rectangle is 64 centimeters and the width is 22 centimeters, what is the perimeter?
- If the length of a rectangle is 32 centimeters and the width is 65.5 centimeters, what is the area?
- If the length of a rectangle is 30 centimeters and the width is 16 centimeters, what is the area?
- If the perimeter of a rectangle is 32 centimeters and the width is 8 centimeters, what is the length?

### Review (Answers)

To see Review answers, open this PDF file and look for section 2.17.

### Additional Resources

Practice: **Applications of Two-Step Equations**