Skip to main content
\(\require{cancel}\newcommand\degree[0]{^{\circ}} \newcommand\Ccancel[2][black]{\renewcommand\CancelColor{\color{#1}}\cancel{#2}} \newcommand{\alert}[1]{\boldsymbol{\color{magenta}{#1}}} \newcommand{\blert}[1]{\boldsymbol{\color{blue}{#1}}} \newcommand{\bluetext}[1]{\color{blue}{#1}} \delimitershortfall-1sp \newcommand\abs[1]{\left|#1\right|} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \)

SectionApplications of Linear Equations

Algebra simplifies the process of solving real-world problems. This is done by using letters to represent unknowns, restating problems in the form of equations, and by offering systematic techniques for solving those equations. To solve problems using algebra, first translate the wording of the problem into mathematical statements that describe the relationships between the given information and the unknowns. Usually, this translation to mathematical statements is the difficult step in the process. The key to the translation is to carefully read the problem and identify certain key words and phrases.

Key Words Translation
Sum: increased by, more than, plus, added to, total \(+\)
Difference: decreased by, subtracted from, less, minus \(-\)
Product: multiplied by, of, times, twice \(\cdot\)
Quotient: divided by, ratio, per \(\div\)
Equals: is, total, result \(=\)

When translating sentences into mathematical statements, be sure to read the sentence several times and parse out the key words and phrases. It is important to first identify the variable, let \(x\) represent it, and state in words what the unknown quantity is. This step not only makes our work more readable, but also forces us to think about what we are looking for.


When 6 is subtracted from twice the sum of a number and 8, the result is 5. Find the number.


Let \(n\) represent the unknown number.

\begin{equation*} 2\cdot (n+8)-6=5 \end{equation*}

To understand why we included the parentheses in the set up, you must study the structure of the following two sentences and their translations:

"twice the sum of a number and \(8\)" \(2(n+8)\)
"the sum of twice a number and \(8\)" \(2n+8\)
The key was to focus on the phrase twice the sum, this prompted us to group the sum within parentheses and then multiply by \(2\text{.}\) After translating the sentence into a mathematical statement we then solve.

\begin{equation*} \begin{aligned} 2(n+8)-6 \amp = 5\\ 2n+16-6\amp = 5\\ 2n+10\amp =5\\ 2n\amp = -5\\ n\amp = -\frac{5}{2} \end{aligned} \end{equation*}


\begin{equation*} \begin{aligned} 2(n+8)-6\amp = 2\left(\alert{-\frac{5}{2}}+8\right)-6\\ \amp = 2\left(\frac{11}{2}\right)-6\\ \amp = 11-6\\ \amp =5 \end{aligned} \end{equation*}

Since our result is 5, we know we have the correct answer.

General guidelines for setting up and solving word problems follow:

Guidelines for Word Problems
  1. Read the problem several times, identify the key words and phrases, and organize the given information.
  2. Identify the variables by assigning a letter or expression to the unknown quantities.
  3. Translate and set up an algebraic equation that models the problem.
  4. Solve the resulting algebraic equation.
  5. Finally, answer the question in sentence form and make sure it makes sense (check it).

All of the equations in the following examples can be set up using only one variable. We can avoid two variables by looking for a relationship between the unknowns.


A rectangle has a perimeter measuring \(92\) meters. The length is \(2\) meters less than \(3\) times the width. Find the dimensions of the rectangle.


The sentence "The length is \(2\) meters less than \(3\) times the width" gives us the relationship between the two important quantities, the length and the width. Let \(w\) represent the width of the rectangle. Since the length is \(2\) meters less than \(3\) times the width, the length must be \(3w-2\text{.}\)

The sentence "A rectangle has a perimeter measuring \(92\) meters" suggests an algebraic set up. The perimeter of a rectangle is always the sum of twice the length and twice the width; in this example, this means that the perimeter is

\begin{equation*} 2w+2(3w-2) \end{equation*}

Since we know that the perimeter is \(92\) meters, we have the following equation:

\begin{equation*} 92=2w+2(3x-2) \end{equation*}

Once you have set up an algebraic equation with one variable, solve for the width, \(w\text{.}\)

\begin{equation*} \begin{aligned} 92\amp =2w+2(3w-2) \amp \text{Distribute}\\ 92\amp =2w+6w-4 \amp\text{Combine like terms}\\ 92\amp =8w-4 \amp\text{Solve for }w\\ 96\amp =8w\\ 12\amp =w\\ \end{aligned} \end{equation*}

Use \(3w-2\) to find the length.

\begin{equation*} 3w-2=3(\alert{12})-2=36-2=34 \end{equation*}

This means that the dimensions of the rectangle are \(12\) meters by \(34\) meters. To check, we make sure the perimeter is actually\(92\) meters.

\begin{equation*} 2(12)+2(34)=24+68=92 \end{equation*}

SubsectionInterest Applications


Given a \(4.375\)% annual interest rate, how long will it take \($2,500\) to yield \($437.50\) in simple interest?


Let \(t\) represent the time needed to earn $\(437.50\) at \(4.375\)%. Organize the information needed to use the formula for simple interest, \(I=prt\text{.}\)

Given interest for the time period \(I=\) $\(437.50\)
Given principal \(p=\) $\(2,500\)
Given rate \(r=4.375\text{%}=0.04375\)

Next, substitute all of the known quantities into the formula and then solve for the only unknown, \(t\text{.}\)

\begin{equation*} \begin{aligned} I\amp = prt\\ \alert{437.50}\amp = \alert{2500}(\alert{0.04375})t\\ 437.50\amp = 109.375t\\ \frac{437.50}{109.375}\amp = \frac{109.375t}{109.375}\\ 4\amp = t \end{aligned} \end{equation*}

It takes \(4\) years for \($2,500\) invested at \(4.375\)% to earn \($437.50\) in simple interest.


Stephanie invested her total savings of \($12,500\) in two accounts earning simple interest. Her mutual fund account earned \(7\)% last year and her CD earned \(4.5\)%. If her total interest for the year was \($670\text{,}\) how much was in each account?


The sum of the money in each account is \($12,500\text{.}\) When a total is involved, a common technique used to avoid two variables is to represent the second unknown as the difference of the total and the first unknown.

Let \(m\) represent the amount invested in the mutual fund.

Since the amount in the two accounts total \($12,500\text{,}\) the amount in the CD must be \(12,500-m\text{.}\)

Interest earned in the mutual fund (in dollars)
\begin{align*} I\amp = prt\\ \amp = m\cdot 0.07\cdot 1\\ \amp =0.07m \end{align*}
Interest earned in the CD (in dollars)
\begin{align*} I\amp = prt\\ \amp = (12,500-m)\cdot 0.045\cdot 1\\ \amp = 0.045(12,500-m) \end{align*}
Total Interest \($670\)

The total interest is the sum of the interest earned from each account.

\begin{equation*} \begin{aligned} \text{mutual fund interest}+\text{CD interest} \amp = \text{total interest}\\ 0.07m +0.045(12,500-m)\amp = 670 \end{aligned} \end{equation*}

This equation models the problem with one variable. Solve for \(m\text{.}\)

\begin{equation*} \begin{aligned} 0.07m+0.045(12,500-m)\amp = 670\\ 0.07m+562.5-0.045m\amp = 670\\ 0.025m+562.5\amp = 670\\ 0.025m\amp = 107.5\\ m\amp = \frac{107.5}{0.025}\\ m\amp =4,300 \end{aligned} \end{equation*}

Use \(12,500-m\) to find the amount in the CD.

\begin{equation*} 12,500-m=12,500-\alert{4300}=8,200 \end{equation*}

Stephanie invested \($4,300\) at \(7\)% in a mutual fund and \($8,200\) at \(4.5\)% in a CD.

SubsectionCurrency Applications

A new kind of problem we come across in applications is something we call currency problems. These problems involve finding the number of coins or bills given other information about value. It is important to be careful in defining our variables specifically - either for the number of currency or the value of the currency.


Michael has \(14\) bills in his wallet altogether. If the wallet contains only $\(5\) and $\(10\) bills and Michael has 3 $\(10\) bills, how much money does Michael have in his wallet?


Let \(f\) be the number of five dollar bills. Because Michael has 14 bills in his wallet, and all are either fives or tens, he must have \(14-3=11\) fives in his wallet.

Since a $\(5\) bill is worth \(5\) dollars, \(5f\) will give us the money amount of all the $\(5\) bills. The portion of the sentence which tells us that Michael has 3 $\(10\) bills tells us that the total amount of money in Michael's wallet is \(5*f+3*10=5f+30\) where \(f\) represents the number of $\(5\) in Michael's wallet. Since we have set up an linear equation, we can plug in the number of $\(5\) bills, 11:

\begin{equation*} 5*11+30=85 \end{equation*}

Michael has nine $\(85\) in his wallet.


Let \(q\) represent the number of quarters. Since Ashley has nine more fifty-cent pieces than quarters, the number of fifty cent pieces she has is \(q+9\text{.}\)

Since a quarter is worth \($0.25\text{,}\) \(0.25\) represents the value of Ashley's quarters. Similarly, since a fifty-cent piece is worth \($0.50\text{,}\) \(0.50(q+9)\) gives the value of Ashley's fifty-cent pieces.

Since Ashley has a total of \($17.25\text{,}\) it must be the case that the sum of these values is \(17.25\text{:}\)

\begin{equation*} 0.25q+0.50(q+9)=17.25 \end{equation*}

We can now solve for \(q\text{:}\)

\begin{equation*} \begin{aligned} 0.25q+0.50(q+9) \amp= 17.25 \\ 0.25q+0.50q+4.50 \amp= 17.25 \\ 0.75q+4.50 \amp= 17.25 \\ 0.75q \amp= 12.75 \\ q \amp= 17 \end{aligned} \end{equation*}

Ashley has \(17\) quarters and \(26\) fifty-cent pieces.

SubsectionMixing Applications

Mixture problems often include a percentage and some total amount. It is important to make a distinction between these two types of quantities. For example, if a problem states that a 10-ounce container is filled with a \(3\)% saline (salt) solution, then this means that the container is filled with a mixture of salt and water as follows:

\(\) \(Percentage\) \(Amount\)
\(Salt\) \(3\)%\(=0.03\) \(0.03\) (\(10\) ounces)\(=0.3\) ounces
\(Water\) \(97\)%\(=0.96\) \(0.96\) (\(10\) ounces)\(=9.7\) ounces

In other words, we multiply the percentage times the total to get the amount of each part of the mixture.


How many milliliters of a \(15\)% alcohol solution must be mixed with \(110\) milliliters of a \(35\)% solution to obtain a \(25\)% solution?


Let \(m\) be the milliliters of the \(15\)% solution.

To get the amount of the first part of the mixture, then, we multiply the percentage by the amount: \(0.15m\text{.}\)

Likewise, to get the amount of the second part of the mixture, we multiply the percentage by the amount: \(0.35\cdot 110=38.5\text{.}\)

Next we look at this problem from the overall viewpoint - we want a \(25\)% solution in the end. So again, to get the amount of the entire mixture, we multiply the percentage by the amount: \(0.25(m+110)\text{.}\)

Now we have all the pieces of our equation, so we can set it up to solve:

\begin{equation*} \begin{aligned} 0.15m+38.5\amp = 0.25(m+110)\\ 0.15m +38.5 \amp =0.25m+27.5\\ 38.5\amp = 0.10m+27.5\\ 11\amp =0.10m\\ 110\amp =m \end{aligned} \end{equation*}

We have found that we must mix \(110\) milliliters of a \(15\)% alcohol solution with \(110\) milliliters of a \(35\)% solution to obtain a \(25\)% solution.

SubsectionDistance Applications

The distance traveled is equal to the average rate times the time traveled at that rate, \(d=r\cdot t\text{.}\) These problems usually have a lot of data, so it helps to carefully rewrite the pertinent information as you define variables.


Two planes leave a city traveling in opposite directions. One travels at a rate of \(530\) mph and the other at a rate of \(600\) mph. How long will it take until they are \(3672.5\) miles apart?


The first plane travels at a rate of \(530\) mph, meaning that its distance is modeled by the equation

\begin{equation*} d_1=r\cdot t= 530 t \end{equation*}

The second plane travels at a rate of \(600\) mph, meaning that its distance is modeled by the equation

\begin{equation*} d_2=r\cdot t= 600 t \end{equation*}

Now since the planes travel in opposite directions, the distance is getting bigger as time goes on, meaning that over time, we can measure the distance between the planes by adding the two distances:

\begin{equation*} d=d_1+d_2=530t+600t=1130t \end{equation*}

Now, we want to know when the distance is \(3672.5\text{.}\) Thus we solve:

\begin{equation*} \begin{aligned} d\amp = 1130t\\ 3672.5\amp =1130t\\ 3.25\amp =t \end{aligned} \end{equation*}

This tells us that it will take \(3\) hours and \(15\) minutes for the planes to be \(3672.5\) miles apart.

Dan and Ariel are \(385.825\) miles apart by sea. Dan sails north on his yacht towards Ariel, while Ariel sails south on her yacht towards Dan. Dan travels at \(11\) knots (\(12.65\) mph), while Ariel travels at \(50\) knots (\(57.5\) mph). How long will it take until they meet?