
1.4: Problem Solving


Learning Objectives

• Define a process for problem solving
• Translate words into algebraic expressions and equations
• Define a process for solving word problems
• Solve problems containing rates
• Apply the steps for solving word problems to distance, rate, and time problems
• Apply the steps for solving word problems to interest rate problems
• Evaluate a formula using substitution
• Rearrange formulas to isolate specific variables
• Identify an unknown given a formula
• Solve additional applications of linear equations
• Apply the steps for solving word problems to geometry problems
• Use the formula for converting between Fahrenheit and Celsius
• Evaluate a formula using substitution
• Rearrange formulas to isolate specific variables
• Identify an unknown given a formula

Define a Process for Problem Solving

Word problems can be tricky. Often it takes a bit of practice to convert an English sentence into a mathematical sentence, which is one of the first steps to solving word problems. In the table below, words or phrases commonly associated with mathematical operators are categorized. Word problems often contain these or similar words, so it’s good to see what mathematical operators are associated with them.

Addition $$-$$ Multiplication $$=$$
More than Less than Double A number Is
Together In the past Product Often, a value for which no information is given. The same as
Sum slower than times After how many hours?
Total the remainder of How much will it cost?
In the future difference
faster than

Some examples follow:

• $$x=5$$
• Three more than a number becomes $$x+3$$
• Four less than a number becomes $$x-4$$
• Double the cost becomes $$2\cdot\text{ cost }$$
• Groceries and gas together for the week cost $250 means $$\text{ groceries }+\text{ gas }=250$$ • The difference of 9 and a number becomes $$9-x$$. Notice how 9 is first in the sentence and the expression Let’s practice translating a few more English phrases into algebraic expressions. Example Translate the table into algebraic expressions:  some number the sum of the number and 3 twice the sum of the number and 3 a length double the length double the length, decreased by 6 a cost the difference of the cost and 20 2 times the difference of the cost and 20 some quantity the difference of 5 and the quantity the difference of 5 and the quantity, divided by 2 an amount of time triple the amount of time triple the amount of time, increased by 5 a distance the sum of $$-4$$ and the twice the distance [reveal-answer q=”790402″]Show Solution[/reveal-answer] [hidden-answer a=”790402″]  $$a+3$$ $$l$$ $$2l-6$$ $$c-20$$ $$q$$ $$\frac{5-q}{2}$$ $$3t$$ $$d$$ $$-4+2d$$ [/hidden-answer] In this example video, we show how to translate more words into mathematical expressions. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=50 The power of algebra is how it can help you model real situations in order to answer questions about them. Here are some steps to translate problem situations into algebraic equations you can solve. Not every word problem fits perfectly into these steps, but they will help you get started. 1. Read and understand the problem. 2. Determine the constants and variables in the problem. 3. Translate words into algebraic expressions and equations. 4. Write an equation to represent the problem. 5. Solve the equation. 6. Check and interpret your answer. Sometimes writing a sentence helps. Example Twenty-eight less than five times a certain number is 232. What is the number? [reveal-answer q=”720402″]Show Solution[/reveal-answer] [hidden-answer a=”720402″] Following the steps provided: 1. Read and understand: we are looking for a number. 2. Constants and variables: 28 and 232 are constants, “a certain number” is our variable because we don’t know its value, and we are asked to find it. We will call it x. 3. Translate: five times a certain number translates to $$5x-28$$ because subtraction is built backward. is 232 translates to $$=232$$ because “is” is associated with equals. 4. Write an equation: $$5x-28=232$$ 5. Solve the equation using what you know about solving linear equations: $$\begin{array}{r}5x-28=232\\5x=260\\x=52\,\,\,\end{array}$$ 6. Check and interpret: We can substitute 52 for x. $$\begin{array}{r}5\left(52\right)-28=232\\5\left(52\right)=260\\260=260\end{array}$$. TRUE! [/hidden-answer] In the video that follows, we show another example of how to translate a sentence into a mathematical expression using a problem solving method. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=50 Another type of number problem involves consecutive numbers. Consecutive numbers are numbers that come one after the other, such as 3, 4, 5. If we are looking for several consecutive numbers it is important to first identify what they look like with variables before we set up the equation. For example, let’s say I want to know the next consecutive integer after 4. In mathematical terms, we would add 1 to 4 to get 5. We can generalize this idea as follows: the consecutive integer of any number, x, is $$x+1$$. If we continue this pattern we can define any number of consecutive integers from any starting point. The following table shows how to describe four consecutive integers using algebraic notation.  First $$x+1$$ Third $$x+3$$ We apply the idea of consecutive integers to solving a word problem in the following example. Example The sum of three consecutive integers is 93. What are the integers? [reveal-answer q=”120402″]Show Solution[/reveal-answer] [hidden-answer a=”120402″] Following the steps provided: 1. Read and understand: We are looking for three numbers, and we know they are consecutive integers. 2. Constants and Variables: 93 is a constant. The first integer we will call x. Second: $$x+2$$ 3. Translate: The sum of three consecutive integers translates to $$=93$$ because is is associated with equals. 4. Write an equation: $$x+\left(x+1\right)+\left(x+2\right)=93$$ 5. Solve the equation using what you know about solving linear equations: We can’t simplify within each set of parentheses, and we don’t need to use the distributive property so we can rewrite the equation without parentheses. $$\begin{array}{r}x+x+1+x+2=93\\3x+3 = 93\\\underline{-3\,\,\,\,\,-3}\\3x=90\\\frac{3x}{3}=\frac{90}{3}\\x=30\end{array}$$ 6. Check and Interpret: Okay, we have found a value for x. We were asked to find the value of three consecutive integers, so we need to do a couple more steps. Remember how we defined our variables: The first integer we will call $$x=30$$ Second: $$30+1=31$$ Third: $$30+2=32$$ The three consecutive integers whose sum is $$30\text{, }31\text{, and }32$$ [/hidden-answer] In the following video we show another example of a consecutive integer problem. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=50 Rates There is often a well-known formula or relationship that applies to a word problem. For example, if you were to plan a road trip, you would want to know how long it would take you to reach your destination. $$d=rt$$ is a well-known relationship that associates distance traveled, the rate at which you travel, and how long the travel takes. Distance, Rate, and Time If you know two of the quantities in the relationship $$30\frac{\text{ miles }}{\text{ hour }}$$ for 2 hours, you can find the distance you would travel by multiplying rate times time or $$\left(30\frac{\text{ miles }}{\text{ hour }}\right)\left(2\text{ hours }\right)=60\text{ miles }$$. We can generalize this idea depending on what information we are given and what we are looking for. For example, if we need to find time, we could solve the $$d=rt$$ equation for t using division: $$d=rt\\\frac{d}{r}=t$$ Likewise, if we want to find rate, we can isolate r using division: $$d=rt\\\frac{d}{t}=r$$ In the following examples you will see how this formula is applied to answer questions about ultra marathon running. Ann Trason Ultra marathon running (defined as anything longer than 26.2 miles) is becoming very popular among women even though it remains a male-dominated niche sport. Ann Trason has broken twenty world records in her career. One such record was the American River 50-mile Endurance Run which begins in Sacramento, California, and ends in Auburn, California.[1] In 1993 Trason finished the run with a time of 6:09:08. The men’s record for the same course was set in 1994 by Tom Johnson who finished the course with a time of 5:33:21.[2] In the next examples we will use the $$d=rt$$ formula to answer the following questions about the two runners. 1. What was each runner’s rate for their record-setting runs? 2. By the time Johnson had finished, how many more miles did Trason have to run? 3. How much further could Johnson have run if he had run as long as Trason? 4. What was each runner’s time for running one mile? To make answering the questions easier, we will round the two runners’ times to 6 hours and 5.5 hours. Example What was each runner’s rate for their record-setting runs? [reveal-answer q=”55589″]Show Solution[/reveal-answer] [hidden-answer a=”55589″] Read and Understand: We are looking for rate and we know distance and time, so we can use the idea: $$d=rt\\\frac{d}{t}=r$$ Define and Translate: Because there are two runners, making a table to organize this information helps. Note how we keep units to help us keep track of what how all the terms are related to each other. Runner Distance = (Rate ) (Time) Trason 50 miles r 6 hours Johnson 50 miles r 5.5 hours Write and Solve: Trason’s rate: $$d=rt\\\frac{d}{t}=r$$ $$\begin{array}{c}d=rt\\\\50\text{ miles }=\text{r}\left(6\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{8.33\text{ miles }}{\text{ hour }}\end{array}$$. (rounded to two decimal places) Johnson’s rate: $$d=rt\\\frac{d}{t}=r$$ $$\begin{array}{c}d=rt\\\\,\,\,\,\,\,\,50\text{ miles }=\text{r}\left(5.5\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{9.1\text{ miles }}{\text{ hour }}\end{array}$$ (rounded to two decimal places) Check and Interpret: We can fill in our table with this information. Runner Distance = (Rate ) (Time) Trason 50 miles 8.33 $$\frac{\text{ miles }}{\text{ hour }}$$ 5.5 hours [/hidden-answer] Now that we know each runner’s rate we can answer the second question. Example By the time Johnson had finished, how many more miles did Trason have to run? [reveal-answer q=”747303″]Show Solution[/reveal-answer] [hidden-answer a=”747303″] Here is the table we created for reference: Runner Distance = (Rate ) (Time) Trason 50 miles 8.33 $$\frac{\text{ miles }}{\text{ hour }}$$ 5.5 hours Read and Understand: We are looking for how many miles Trason still had on the trail when Johnson had finished after 5.5 hours. This is a distance, and we know rate and time. Define and Translate: We can use the formula $$d=rt$$ again. This time the unknown is d, and the time Trason had run is 5.5 hours. Write and Solve: $$\begin{array}{l}d=rt\\\\d=8.33\frac{\text{ miles }}{\text{ hour }}\left(5.5\text{ hours }\right)\\\\d=45.82\text{ miles }\end{array}$$. Check and Interpret: Have we answered the question? We were asked to find how many more miles she had to run after 5.5 hours. What we have found is how long she had run after 5.5 hours. We need to subtract $$d=45.82\text{ miles }$$ from the total distance of the course. $$50\text{ miles }-45.82\text{ miles }=1.48\text{ miles }$$ [/hidden-answer] The third question is similar to the second. Now that we know each runner’s rate, we can answer questions about individual distances or times. Examples How much further could Johnson have run if he had run as long as Trason? [reveal-answer q=”757303″]Show Solution[/reveal-answer] [hidden-answer a=”757303″] Here is the table we created for reference: Runner Distance = (Rate ) (Time) Trason 50 miles 8.33 $$\frac{\text{ miles }}{\text{ hour }}$$ 5.5 hours Read and Understand: The word further implies we are looking for a distance. Define and Translate: We can use the formula $$9.1\frac{\text{ miles }}{\text{ hour }}$$ Write and Solve: $$\begin{array}{l}d=rt\\\\d=9.1\frac{\text{ miles }}{\text{ hour }}\left(6\text{ hours }\right)\\\\d=54.6\text{ miles }\end{array}$$. Check and Interpret: Have we answered the question? We were asked to find how many more miles Johnson would have run if he had run at his rate of $$9.1\frac{\text{ miles }}{\text{ hour }}$$ for 6 hours. Johnson would have run 54.6 miles, so that’s 4.6 more miles than than he ran for the race. [/hidden-answer] Now we will tackle the last question where we are asked to find a time for each runner. Example What was each runner’s time for running one mile? [reveal-answer q=”757309″]Show Solution[/reveal-answer] [hidden-answer a=”757309″] Here is the table we created for reference: Runner Distance = (Rate ) (Time) Trason 50 miles 8.33 $$\frac{\text{ miles }}{\text{ hour }}$$ 5.5 hours Read and Understand: we are looking for time, and this time our distance has changed from 50 miles to 1 mile, so we can use $$d=rt\\\frac{d}{r}=t$$ Define and Translate: we can use the formula $$d=rt$$ again. This time the unknown is t, the distance is 1 mile, and we know each runner’s rate. It may help to create a new table: Runner Distance = (Rate ) (Time) Trason 1 mile 8.33 $$\frac{\text{ miles }}{\text{ hour }}$$ t hours Write and Solve: Trason: We will need to divide to isolate time. $$\begin{array}{c}d=rt\\\\1\text{ mile }=8.33\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{8.33\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.12\text{ hours }=t\end{array}$$. 0.12 hours is about 7.2 minutes, so Trason’s time for running one mile was about 7.2 minutes. WOW! She did that for 6 hours! Johnson: We will need to divide to isolate time. $$\begin{array}{c}d=rt\\\\1\text{ mile }=9.1\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{9.1\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.11\text{ hours }=t\end{array}$$. 0.11 hours is about 6.6 minutes, so Johnson’s time for running one mile was about 6.6 minutes. WOW! He did that for 5.5 hours! Check and Interpret: Have we answered the question? We were asked to find how long it took each runner to run one mile given the rate at which they ran the whole 50-mile course. Yes, we answered our question. Trason’s mile time was $$6.6\frac{\text{minutes}}{\text{mile}}$$ [/hidden-answer] In the following video, we show another example of answering many rate questions given distance and time. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=50 Simple Interest In order to entice customers to invest their money, many banks will offer interest-bearing accounts. The accounts work like this: a customer deposits a certain amount of money (called the Principal, or P), which then grows slowly according to the interest rate (R, measured in percent) and the length of time (T, usually measured in months) that the money stays in the account. The amount earned over time is called the interest (I), which is then given to the customer. Beware! Interest rates are commonly given as yearly rates, but can also be monthly, quarterly, bimonthly, or even some custom amount of time. It is important that the units of time and the units of the interest rate match. You will see why this matters in a later example. The simplest way to calculate interest earned on an account is through the formula $$\displaystyle I=P\,\cdot \,R\,\cdot \,T$$. If we know any of the three amounts related to this equation, we can find the fourth. For example, if we want to find the time it will take to accrue a specific amount of interest, we can solve for T using division: $$\displaystyle\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,I=P\,\cdot \,R\,\cdot \,T\\\\ \frac{I}{{P}\,\cdot \,R}=\frac{P\cdot\,R\,\cdot \,T}{\,P\,\cdot \,R}\\\\\,\,\,\,\,\,\,\,\,\,\,{T}=\frac{I}{\,R\,\cdot \,T}\end{array}$$ Below is a table showing the result of solving for each individual variable in the formula. Solve For Result I $${P}=\frac{I}{{R}\,\cdot \,T}$$ R $${T}=\frac{I}{{P}\,\cdot \,R}$$ In the next examples, we will show how to substitute given values into the simple interest formula, and decipher which variable to solve for. Example If a customer deposits a principal of$2000 at a monthly rate of 0.7%, what is the total amount that she has after 24 months?

Substitute in the values given for the Principal, Rate, and Time.

$$\displaystyle\begin{array}{l}I=P\,\cdot \,R\,\cdot \,T\\I=2000\cdot 0.7\%\cdot 24\end{array}$$

Rewrite 0.7% as the decimal 0.007, then multiply.

$$\begin{array}{l}I=2000\cdot 0.007\cdot 24\\I=336\end{array}$$

Add the interest and the original principal amount to get the total amount in her account.

$$\displaystyle 2000+336=2336$$

She has $2336 after 24 months.[/hidden-answer] The following video shows another example of finding an account balance after a given amount of time, principal invested, and a rate. A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=50 In the following example you will see why it is important to make sure the units of the interest rate match the units of time when using the simple interest formula. Example Alex invests$600 at 3.25% monthly interest for 3 years. What amount of interest has Alex earned?

Read and Understand: The question asks for an amount, so we can substitute what we are given into the simple interest formula $$I=P\,\cdot \,R\,\cdot \,T$$

Define and Translate: we know P, R, and T so we can use substitution. R = 0.0325, P = $600, and T = 3 years. We have to be careful! R is in months, and T is in years. We need to change T into months because we can’t change the rate—it is set by the bank. $${T}=3\text{ years }\cdot12\frac{\text{ months }}{ year }=36\text{ months }$$ Write and Solve: Substitute the given values into the formula. $$\begin{array}{l} I=P\,\cdot \,R\,\cdot \,T\\\\I=600\,\cdot \,0.035\,\cdot \,36\\\\{I}=756\end{array}$$ Check and Interpret: We were asked what amount Alex earned, which is the amount provided by the formula. In the previous example we were asked the total amount in the account, which included the principal and interest earned. Alex has earned$756.

In the following video we show another example of how to find the amount of interest earned after an investment has been sitting for a given monthly interest.

A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=50

Example

After 10 years, Jodi’s account balance has earned $1080 in interest. The rate on the account is 0.09% monthly. What was the original amount she invested in the account? [reveal-answer q=”97641″]Show Solution[/reveal-answer] [hidden-answer a=”97641″] Read and Understand: The question asks for the original amount invested, the principal. We are given a length of time in years, and an interest rate in months, and the amount of interest earned. Define and Translate: we know I =$1080, R = 0.009, and T = 10 years so we can use $${P}=\frac{I}{{R}\,\cdot \,T}$$

We also need to make sure the units on the interest rate and the length of time match, and they do not. We need to change time into months again.

$${T}=10\text{ years }\cdot12\frac{\text{ months }}{ year }=120\text{ months }$$

Write and Solve:

Substitute the given values into the formula

$$\begin{array}{l}{P}=\frac{I}{{R}\,\cdot \,T}\\\\{P}=\frac{1080}{{0.009}\,\cdot \,120}\\\\{P}=\frac{1080}{1.08}=1000\end{array}$$

Check and Interpret:

We were asked to find the principal given the amount of interest earned on an account. If we substitute P = $1000 into the formula $$I=P\,\cdot \,R\,\cdot \,T$$ we get $$I=1000\,\cdot \,0.009\,\cdot \,120\\I=1080$$ Our solution checks out. Jodi invested$1000.

The last video shows another example of finding the principle amount invested based on simple interest.

A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=50

In the next section we will apply our problem-solving method to problems involving dimensions of geometric shapes.

Further Applications of Linear Equations

Formulas come up in many different areas of life. We have seen the formula that relates distance, rate, and time and the formula for simple interest on an investment. In this section we will look further at formulas and see examples of formulas for dimensions of geometric shapes as well as the formula for converting temperature between Fahrenheit and Celsius.

Geometry

There are many geometric shapes that have been well studied over the years. We know quite a bit about circles, rectangles, and triangles. Mathematicians have proven many formulas that describe the dimensions of geometric shapes including area, perimeter, surface area, and volume.

Perimeter

Perimeter is the distance around an object. For example, consider a rectangle with a length of 8 and a width of 3. There are two lengths and two widths in a rectangle (opposite sides), so we add $${P}=2\left({L}\right)+2\left({W}\right)$$ where

L = Length

W = Width

In the following example, we will use the problem-solving method we developed to find an unknown width using the formula for the perimeter of a rectangle. By substituting the dimensions we know into the formula, we will be able to isolate the unknown width and find our solution.

Example

You want to make another garden box the same size as the one you already have. You write down the dimensions of the box and go to the lumber store to buy some boards. When you get there you realize you didn’t write down the width dimension—only the perimeter and length. You want the exact dimensions so you can have the store cut the lumber for you.

Here is what you have written down:

Perimeter = 16.4 feet
Length = 4.7 feet

Can you find the dimensions you need to have your boards cut at the lumber store? If so, how many boards do you need and what lengths should they be?

Read and Understand: We know perimeter = 16.4 feet and length = 4.7 feet, and we want to find width.

Define and Translate:

Define the known and unknown dimensions:

w = width

p = 16.4

l = 4.7

Write and Solve:

First we will substitute the dimensions we know into the formula for perimeter:

$$\begin{array}{l}\,\,\,\,\,P=2{W}+2{L}\\\\16.4=2\left(w\right)+2\left(4.7\right)\end{array}$$

Then we will isolate w to find the unknown width.

$$\begin{array}{l}16.4=2\left(w\right)+2\left(4.7\right)\\16.4=2{w}+9.4\\\underline{-9.4\,\,\,\,\,\,\,\,\,\,\,\,\,-9.4}\\\,\,\,\,\,\,\,7=2\left(w\right)\\\,\,\,\,\,\,\,\frac{7}{2}=\frac{2\left(w\right)}{2}\\\,\,\,\,3.5=w\end{array}$$

Write the width as a decimal to make cutting the boards easier and replace the units on the measurement, or you won’t get the right size of board!

Check and Interpret:

If we replace the width we found, $$w=3.5\text{ feet }$$ into the formula for perimeter with the dimensions we wrote down, we can check our work:

$$\begin{array}{l}\,\,\,\,\,{P}=2\left({L}\right)+2\left({W}\right)\\\\{16.4}=2\left({4.7}\right)+2\left({3.5}\right)\\\\{16.4}=9.4+7\\\\{16.4}=16.4\end{array}$$

Our calculation for width checks out. We need to ask for 2 boards cut to 3.5 feet and 2 boards cut to 4.7 feet so we can make the new garden box.

This video shows a similar garden box problem.

A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=50

We could have isolated the w in the formula for perimeter before we solved the equation, and if we were going to use the formula many times, it could save a lot of time. The next example shows how to isolate a variable in a formula before substituting known dimensions or values into the formula.

Example

Isolate the term containing the variable, w, from the formula for the perimeter of a rectangle:

$${P}=2\left({L}\right)+2\left({W}\right)$$.

First, isolate the term with w by subtracting 2l from both sides of the equation.

$$\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,p\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\p-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}$$

Next, clear the coefficient of w by dividing both sides of the equation by 2.

$$\displaystyle \begin{array}{l}\underline{p-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\frac{p-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\frac{p-2l}{2}\end{array}$$

You can rewrite the equation so the isolated variable is on the left side.

$$w=\frac{p-2l}{2}$$

Area

The area of a triangle is given by $$A=\frac{1}{2}bh$$ where

A = area
b = the length of the base
h = the height of the triangle

Remember that when two variables or a number and a variable are sitting next to each other without a mathematical operator between them, you can assume they are being multiplied. This can seem frustrating, but you can think of it like mathematical slang. Over the years, people who use math frequently have just made that shortcut enough that it has been adopted as convention.

In the next example we will use the formula for area of a triangle to find a missing dimension, as well as use substitution to solve for the base of a triangle given the area and height.

Example

Find the base (b) of a triangle with an area (A) of 20 square feet and a height (h) of 8 feet.

Use the formula for the area of a triangle, $${A}=\frac12{bh}$$.

Substitute the given lengths into the formula and solve for b.

$$\displaystyle \begin{array}{l}\,\,A=\frac{1}{2}bh\\\\20=\frac{1}{2}b\cdot 8\\\\20=\frac{8}{2}b\\\\20=4b\\\\\frac{20}{4}=\frac{4b}{4}\\\\ \,\,\,5=b\end{array}$$

The base of the triangle measures 5 feet.[/hidden-answer]

We can rewrite the formula in terms of b or h as we did with perimeter previously. This probably seems abstract, but it can help you develop your equation-solving skills, as well as help you get more comfortable with working with all kinds of variables, not just x.

Example

Use the multiplication and division properties of equality to isolate the variable b.

$$\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\frac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}$$

Write the equation with the desired variable on the left-hand side as a matter of convention:

$$b=\frac{2A}{h}$$

Use the multiplication and division properties of equality to isolate the variable h.

$$\begin{array}{l}\,\,\,\,\,\,\,\,A=\frac{1}{2}bh\\\\\left(2\right)A=\left(2\right)\frac{1}{2}bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\frac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\frac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}$$

Write the equation with the desired variable on the left-hand side as a matter of convention:

$$h=\frac{2A}{b}$$

The following video shows another example of finding the base of a triangle given area and height.

A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=50

Temperature

Let’s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.

$$C=\left(F--32\right)\cdot \frac{5}{9}$$

Example

Given a temperature of $${}^{\circ}{F}$$.

Substitute the given temperature in$${}^{\circ}{C}$$ into the conversion formula:

$$12=\left(F-32\right)\cdot \frac{5}{9}$$

Isolate the variable F to obtain the equivalent temperature.

$$\begin{array}{r}12=\left(F-32\right)\cdot \frac{5}{9}\\\\\left(\frac{9}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\frac{108}{5}\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}$$

As with the other formulas we have worked with, we could have isolated the variable F first, then substituted in the given temperature in Celsius.

Example

Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F.

$$C=\left(F--32\right)\cdot \frac{5}{9}$$

To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by $$\displaystyle \frac{9}{5}$$.

$$\begin{array}{l}\\\,\,\,\,\left(\frac{9}{5}\right)C=\left(F-32\right)\left(\frac{5}{9}\right)\left(\frac{9}{5}\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\frac{9}{5}C=F-32\end{array}$$

$$\begin{array}{l}\frac{9}{5}\,C+32=F-32+32\\\\\frac{9}{5}\,C+32=F\end{array}$$

$$F=\frac{9}{5}C+32$$[/hidden-answer]

Express the formula for the surface area of a cylinder, $$s=2\pi rh+2\pi r^{2}$$, in terms of the height, h.

In this example, the variable h is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to write down what you think is the best first step to take to isolate h.

[practice-area rows=”1″][/practice-area]
Isolate the term containing the variable, h, by subtracting $$2\pi r^{2}$$from both sides.

$$\begin{array}{r}S\,\,=2\pi rh+2\pi r^{2} \\ \underline{-2\pi r^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,-2\pi r^{2}}\\S-2\pi r^{2}\,\,\,\,=\,\,\,\,2\pi rh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}$$

Next, isolate the variable h by dividing both sides of the equation by $$2\pi r$$.

$$\begin{array}{r}\frac{S-2\pi r^{2}}{2\pi r}=\frac{2\pi rh}{2\pi r} \\\\ \frac{S-2\pi r^{2}}{2\pi r}=h\,\,\,\,\,\,\,\,\,\,\end{array}$$

You can rewrite the equation so the isolated variable is on the left side.

$$h=\frac{S-2\pi r^{2}}{2\pi r}$$[/hidden-answer]

In the last video, we show how to convert from celsius to fahrenheit.

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1. "Ann Trason." Wikipedia. Accessed May 05, 2016. https://en.Wikipedia.org/wiki/Ann_Trason. ↵
2. "American River 50 Mile Endurance Run." Wikipedia. Accessed May 05, 2016. https://en.Wikipedia.org/wiki/American_River_50_Mile_Endurance_Run. ↵