Setting up equations

Often times the most difficult part of solving a word problem is figuring out how to set up the correct equation from the words. Instead of being giving a clear equation to solve like this:

You’re given a paragraph like this:

Janetta wants to start saving money to buy a house. She needs to spend $1,300 a month to cover all of her basic living expenses (food, rent, utilities, insurance). She plans to save 80% of all the money that she doesn’t need to spend on her basic living expenses. If Janet earns $3,000 a month, how much will she save in a year?

The equation for solving Janetta’s annual savings is the one we gave above. \( (3,000-1,200) \) is the amount of money Janetta will have left over each month after covering her basic living expenses. \( 0.80 \) is the percentage of this left over money that Janetta wants to save. And, finally since \( 12 \) is the number of months in a year, we multiply these three expressions together to get the amount of money Janetta will save in a year, which we call \( S \):

We could have omitted the exact value of Janetta’s monthly income out and just called this \( I \). If we did this and asked you to set up the equation for her annual savings you would have gotten this:

We could have also left out the exact savings rate Janetta wanted to pursue and just called the rate \( r \), and also just called her living expenses \( E \). This would have led to the equation:

It is not possible to solve either of these equations because there are too many missing variables, but that’s okay. Some test questions will only ask you to set up an equation like these and will not require you to solve for any exact values. Setting up generic equations like these can be helpful in modeling situations. In this case we could alter values of \( r \), \( I \), and \( E \) to check how the outcome (\( S \)) changes. For example if \( r=0.50 \), \( I=4,000 \), and \( E=2,000 \) then \( S \) would be:

Let’s turn to another example: A study of the weight of American children between \( 1 \) and \( 10 \) years old predicted that a child’s weight in kilograms should be equal to ten more than twice their age in years. Matt is \( n \) years old. Write an equation that gives Matt’s predicted weight (\( w \)).

We must pull the information out of the above paragraph that mathematically relates age to predicted weight. That information is given here: “a child’s weight in kilograms should be equal to ten more than twice their age in years.” Rewriting this in mathematical terms using \( n \) to denote “age in years,” and \( w \) to denote “weight in kilograms,” we get:

The question only asked for the equation so we are done. If we were given a value of \( n \) we could substitute it into the equation and solve for \( w \), but that is not necessary here.

Some questions may require a few steps to solve. The following is an example of one:

Thomas has two more apples than Uriah. Uriah has three times as many apples as Victor. If Victor has three apples, how many apples does Thomas have?

To begin solving such a problem we’ll give all of the variables short names; we’ll let the number of apples Thomas has be represented by the letter \( T \), and define \( U \) and \( V \) similarly. The problem gives us relationships between \( T \), \( U \), and \( V \). Expressing these relationships mathematically should be our next step. The words: “Thomas has two more apples than Uriah” equate to the equation:

The words: “Uriah has three times as many apples as Victor,” equate to the equation:

And finally, we’re given that “Victor has 3 apples,” which means \( V=3 \). Now we’re ready to solve for \( T \). We can do so by substituting \( V=3 \) into the second equation to solve for \( U \):

Now we’re able to solve for \( T \) by substituting \( U=9 \) into the first equation:

With a few steps we have solved that Thomas has 11 apples. None of the computations we performed were very difficult; the challenge of the problem was purely in setting up the equations we needed to solve and knowing to perform two substitutions. The approach of naming variables and translating words into equations is at the heart of solving word problems. Practice your translation skills with these problems.

Practice questions

1. Candice wants to start saving money to buy a car. She decides to save $200 each month plus an additional 20% of her disposable income. If Candice’s monthly disposable income is I, what will her monthly savings (S) be?
2. Jim decides to save 5% of his monthly income for retirement. His company matches employee savings up to 3% of their incomes. If Jim’s monthly income is Y, how much will he be saving (S) in his retirement account each month?
3. An airplane travelled for 2 hours at a speed of x miles per hour (mph) and 3 hours at a speed 300 mph. The average speed of the airplane’s entire flight was 320 mph. Write an equation that could be used to solve for x.
4. Three cars A, B, and C all drive between Chico, CA and Boise, ID. Car A travels at an average speed that is 10 mph faster than the average speed of car B. Car C travels at an average speed that is 80% of the average speed of car A. Write the average speed of car C in terms of the average speed of car B.
5. Mark is a consultant. He bills clients $200 an hour for the first ten hours, and $100 an hour for each additional hour beyond that. Mark bills Client X for x hours of consulting. If x is greater than 10, what dollar amount (d) does Mark bill Client X for?

Solutions

1. \( S=200+0.2(I) \)
2. \( S=(0.05+0.03)Y = 0.08 \cdot Y \)
3. To derive the equation we need to use the definition of an average. The average speed of this flight is equal to: $$ \frac{x+x+300+300+300}{5}=\frac{2x+3(300)}{5} $$ We can set the above expression equal to the value of the average given to us in the problem to get a solvable equation: $$ \frac{2x+3(300)}{5}=320 $$ This question only asked for us to set up the equation, so we can stop here without taking the additional algebra steps to solve for \( x \)
4. We can pull the following two equations from the word problem: $$ A=B+10 $$ $$ C=0.8A $$ If we substitute the first equation into the second equation we get the average speed of Car C (\( C \)) in terms of the average speed of car B (\( B \)): $$ C=0.8(B+10) $$
5. Mark bills client \( x \) for \( 10 \) hours at the $200/hr rate and (\( x-10 \)) hours at the $100/hr rate, so the total bill is: $$ d=10(200)+(x-10)(100)=2000+100x-1000=100x+1000 $$