Lesson Objectives
  • Demonstrate an understanding of how to translate phrases into algebraic expressions and equations
  • Learn the six-step method used for solving applications of linear equations
  • Learn how to solve word problems that involve "sums of unknown quantities"
  • Learn how to solve word problems that involve the simple interest formula "I = prt"
  • Learn how to solve word problems that involve percentages

How to Solve Word Problems with Linear Equations


In our Algebra 1 course, we learned the six-step method for solving a word problem that involves a linear equation in one variable. Let's quickly review this procedure and then jump into some sample word problems.

Six-step method for Solving Word Problems with Linear Equations in One Variable

  1. Read the problem and determine what you are asked to find
  2. Assign a variable to represent the unknown
    • If more than one unknown exists, we express the other unknowns in terms of this variable
  3. Write out an equation which describes the given situation
  4. Solve the equation
  5. State the answer using a nice clear sentence
  6. Check the result
    • We need to make sure the answer is reasonable. In other words, if asked how many students were on a bus, the answer shouldn't be (-4) as we can't have a negative amount of students on a bus.
Fortunately, pretty much all common Algebra word problems can be broken down into a "type" of problem. The first type of word problem that we will look at involves finding the individual amounts when the sums are known. This problem type is known as "sums of unknown quantities" or "finding unknown quantities".

Finding Unknown Numerical Quantities - Word Problem

Let's look at an example.
Example 1: Solve each word problem
Jay’s Place is a local restaurant that has only three items on its menu. Their menu consists of lasagna, tacos with chips, or steak and rice. One Sunday, the restaurant sold a total of 80 plates from the menu. They sold twice as many plates of tacos with chips as plates of lasagna. Additionally, the sales of plates with Steak and rice were only one-fifth of the sales of plates of lasagna.
Step 1) After reading the problem, it is clear we need to find the number of plates sold for each menu item: lasagna, tacos with chips, and steak with rice.
Step 2) Here we have three unknowns. When this happens, look for the unknown that is involved in both comparisons. This will make our problem much easier to solve.
Since lasagna is involved in each comparison, let's let x be equal to the number of plates of lasagna sold.
x = plates of lasagna
We are told that the restaurant sold twice the number of plates of tacos with chips as plates of lasagna. Since x has been assigned to represent the number of plates of lasagna sold, we can think of twice this number as 2x. This leads us to the following:
then: 2x = plates of tacos with chips
Lastly, we are told that the sales for plates of steak with rice were one-fifth the number of sales of plates of lasagna. Again, since x has been assigned to represent the number of plates of lasagna sold, we can think of one-fifth this number as (1/5)x. This leads us to the following:
then: (1/5)x = plates of steak with rice
Step 3) To write an equation, think about what we know. So far, we have represented how much of each menu item was sold using the variable x. Additionally, we know the total number of plates sold for the day (80). This means we can sum the individual amounts and set this equal to 80 (total plates sold for the day). $$\frac{1}{5}x + x + 2x = 80$$ Step 4) Solve the equation: $$\frac{1}{5}x + x + 2x = 80$$ Multiply both sides by 5, this will clear the fraction: $$x + 5x + 10x = 400$$ $$16x = 400$$ $$x = 25$$ Step 5) Since x represented the number of plates of lasagna sold, we know the following:
25 plates of lasagna
50 plates of tacos with chips (2 x 25 = 50)
5 plates of steak with rice (1/5 x 25 = 5)
Jay's Place sold 25 plates of lasagna, 50 plates of tacos with chips, and 5 plates of steak with rice.
Step 6) We can read back through the problem to check our answer. First, we can check the total number of plates sold:
25 + 50 + 5 = 80
80 = 80
Additionally, we can check that the number of plates sold of tacos with chips are twice that of plates of lasagna:
50 = 2 • 25
50 = 50
Lastly, we can check that the number of plates sold of steak with rice are one-fifth that of plates of lasagna:
25 * 1/5 = 5
5 = 5

Solving Simple Interest Word Problems » I = prt

When we work with investment problems in Algebra, we will encounter two types of interest: simple interest and compound interest. The main difference between the two comes from the fact that compound interest will earn interest on interest. This means as the account balance grows, interest is paid on the current account balance, not the original amount invested (known as the principal). When we work with simple interest word problems, interest is only earned on the principal or amount initially invested. In other words, when we have simple interest, we do not earn interest on interest. Although this makes our calculation much simpler, it is not often used in real life.
With simple interest word problems, we will use the simple interest formula:
I = prt
I » simple interest earned
p » principal or amount invested
r » rate (interest rate as a decimal)
t » time (normally in years, but could be months, days, or whatever given time frame)
Let's look at an example.
Example 2: Solve each word problem
After speaking with a retirement professional, James decides to invest in two different accounts. A bond fund, which pays 5% annual simple interest and a savings account which pays 2% annual simple interest. James first decides to invest $29,000 into the bond fund. If his goal is to make 3% annual simple interest, how much additional money should he invest in the savings account?
Step 1) After reading the problem, it is clear that we need to find the amount that James needs to invest in the savings account in order to have a combined interest rate of 3%.
Step 2) We have one unknown, the amount that is to be invested in the savings account.
let x = amount of money in dollars that will be invested in the savings account
Step 3) To write an equation, let's think about what we know.
James wants an overall interest rate of 3% for the two investments. Currently, we know James will invest $29,000 in the bond fund and earn 5% annual simple interest. If we plug this into the simple interest formula:
I = 29,000(.05)(1)
I = 1450
This amount will be added to the 2% interest obtained from investing an unknown amount (x) in a savings account:
Using our simple interest formula for the savings account yields:
I = x (.02)(1)
I = .02x
If we sum the two, we think about the simple interest earned from the two investments:
I = 1450 + .02x
Since we want I or the simple interest earned to be 3% of the principal invested, we can set up the following equation:
.03(x + 29,000) = .02x + 1450
In other words, 3% multiplied by the principal invested (x + 29,000) will be equal to the interest earned from the savings account (.02x) and the interest earned from the bond fund (1450).
Step 4) Solve the equation:
.03(x + 29,000) = .02x + 1450
.03x + 870 = .02x + 1450
We can clear the decimals by multiplying both side of the equation by 100:
3x + 87,000 = 2x + 145,000
3x - 2x = 145,000 - 87,000
x = 58,000
Step 5) Since x represented the amount that needed to be invested in the savings account, we can state our answer as:
James needs to invest $58,000 dollars in the savings account in order to achieve a simple interest rate of 3% from the two investments.
Step 6) We can read back through the problem to check our answer. Think about the simple interest earned from his two investments:
I = .02(58,000) + .05(29,000)
I = 1160 + 1450
I = 2610
The simple interest earned from the two investments is 2610. If we solve the simple interest formula for r (rate): $$I = prt$$ $$r = \frac{I}{pt}$$ Since time is 1, we can rewrite our equation as: $$r = \frac{I}{p}$$ Since his goal is 3% interest, let's plug everything in and check: $$.03 = \frac{2610}{29,\hspace{-.1em}000 + 58,\hspace{-.1em}000}$$ $$.03 = \frac{2610}{87,\hspace{-.1em}000}$$ .03 = .03

Solving Percent Word Problems

Another common type of word problem involves percentages. These usually deal with percent increase or percent decrease. Let's look at an example.
Example 3: Solve each word problem
Heather works as a cashier for a local grocery store. After one busy weekend shift, she had a total of $2725 in total receipts. This amount included the 9% state and local sales taxes. What was the amount of the tax collected?
Step 1) After reading the problem, it is clear that we need to find the amount of taxes that were collected.
Step 2) We have two unknowns, the pre-tax amount of goods and services and the amount of tax collected.
let x = amount of pre-tax goods and services sold
then .09x = amount of tax collected
Step 3) Write an equation, let's think about what we know.
If we were to sum the amount of pre-tax goods and services sold (x) with the amount of the tax collected (.09x), we would get 2725 (total receipts):
x + .09x = 2725
Step 4) Solve the equation:
x + .09x = 2725
We can multiply both sides by 100 and clear the decimal: 100x + 9x = 272,500
109x = 272,500
x = 2500
Step 5) Since x (2500) represents the amount of pre-tax goods and services sold, we have to subtract this away from the total receipts (2725) to get the amount of the tax collected:
2725 - 2500 = 225
We can state our answer as:
Heather collected $225 in tax during her shift.
Step 6) We can read back through the problem to check our answer. We want to check that the amount of pre-tax goods and services (2500) plus the amount of the tax collected (225) is equal to the total receipts (2725):
2500 + 225 = 2725
2725 = 2725