Most of the time when someone says “word problems” there is automatic panic. But word problems do not have to be the worst part of a math class. By setting up a system and following it, you can be successful with word problems. So what should you do? Here are some recommended steps:  Read the problem carefully and figure out what it is asking you to find.
Usually, but not always, you can find this information at the end of the problem.
 Assign a variable to the quantity you are trying to find.
Most people choose to use x, but feel free to use any variable you like. For example, if you are being asked to find a number, some students like to use the variable n. It is your choice.
 Write down what the variable represents.
At the time you decide what the variable will represent, you may think there is no need to write that down in words. However, by the time you read the problem several more times and solve the equation, it is easy to forget where you started.
 Reread the problem and write an equation for the quantities given in the problem.
This is where most students feel they have the most trouble. The only way to truly master this step is through lots of practice. Be prepared to do a lot of problems.
 Solve the equation.
The examples done in this lesson will be linear equations. Solutions will be shown, but may not be as detailed as you would like. If you need to see additional examples of linear equations worked out completely, click here. (link to linear equations solving.doc)
 Answer the question in the problem.
Just because you found an answer to your equation does not necessarily mean you are finished with the problem. Many times you will need to take the answer you get from the equation and use it in some other way to answer the question originally given in the problem.
 Check your solution.
Your answer should not only make sense logically, but it should also make the equation true. If you are asked for a time value and end up with a negative number, this should indicate that you’ve made an error somewhere. If you are asked how fast a person is running and give an answer of 700 miles per hour, again you should be worried that there is an error. If you substitute these unreasonable answers into the equation you used in step 4 and it makes the equation true, then you should rethink the validity of your equation. Let's Practice: When 6 is added to four times a number, the result is 50. Find the number.
Step 1: What are we trying to find? A number. Step 2: Assign a variable for the number. Let’s call it n. Step 3: Write down what the variable represents. Let n = a number Step 4: Write an equation. We are told 6 is added to 4 times a number. Since n represents the number, four times the number would be 4n. If 6 is added to that, we get . We know that answer is 50, so now we have an equation Step 5: Solve the equation. Step 6: Answer the question in the problem The problem asks us to find a number. We decided that n would be the number, so we have n = 11. The number we are looking for is 11. Step 7: Check the answer. The answer makes sense and checks in our equation from Step 4.
 The sum of a number and 9 is multiplied by 2 and the answer is 8. Find the number.
Step 1: What are we trying to find? A number. Step 2: Assign a variable for the number. Let’s call it n. Step 3: Write down what the variable represents. Let n = a number Step 4: Write an equation. We know that we have the sum of a number and 9 which will give us n + 9. We are then told to multiply that by 2, so we have . Be very careful with your parentheses here. The way this is worded indicates that we find the sum first and then multiply. We also know the answer is 8. So we will solve Step 5: Solve the equation. Step 6: Answer the question in the problem The problem asks us to find a number. We decided that n would be the number, so we have n = 5. The number we are looking for is 5. Step 7: Check the answer. The answer makes sense and checks in our equation from Step 4.
 On an algebra test, the highest grade was 42 points higher than the lowest grade. The sum of the two grades was 138. Find the lowest grade.
Step 1: What are we trying to find? The lowest grade on an algebra test. Step 2: Assign a variable for the lowest test grade. Let’s call it l. Step 3: Write down what the variable represents. Let l = the lowest grade Step 4: Write an equation. Whatever the lowest grade is, we are told that the highest grade is 42 points higher than that. That means we need to add 42 to the lowest grade. This tells us the highest grade is . We also know that the highest grade added to the lowest grade is 138. So, (highest grade) + (lowest grade) = 142. In terms of our variable, Step 5: Solve the equation. Step 6: Answer the question in the problem The problem asks us to find the lowest grade. We decided that l would be the number, so we have l = 48. The lowest grade on the algebra test was 48. Step 7: Check the answer. The answer makes sense and checks in our equation from Step 4.
 At the end of the day, a pharmacist counted and found she has 4/3 as many prescriptions for antibiotics as she did for tranquilizers. She had 84 prescriptions for the two types of drugs. How many prescriptions did she have for tranquilizers?
Step 1: What are we trying to find? The number of prescriptions for tranquilizers. Step 2: Assign a variable for the number of tranquilizer prescriptions. Let’s call it t. Step 3: Write down what the variable represents. Let t = number of tranquilizer prescriptions Step 4: Write an equation. We have to be careful here. The pharmacist had 4/3 as many prescriptions for antibiotics as she did for tranquilizers. Let’s think about this in terms of numbers first. Suppose there were 3 tranquilizer prescriptions, 4/3 as many would mean there were 4 prescriptions for antibiotics. Or if there were 30 tranquilizer prescriptions, then 4/3 as many for antibiotics, would mean there were 40 antibiotic prescriptions. In each case, we are taking the number of tranquilizers and multiplying by 4/3 to get the number of antibiotic prescriptions. So if t is the number of tranquilizer prescriptions, then is the number of antibiotic prescriptions. We are told that together the two types of prescriptions add up to 84. So we end up with the equation . Step 5: Solve the equation. Step 6: Answer the question in the problem The problem asks us to find the number of prescriptions for tranquilizers. We decided that t would be the number of prescriptions for tranquilizers, so we have t = 36. There were 36 prescriptions for tranquilizers. Step 7: Check the answer. The answer makes sense and checks in our equation from Step 4.
 In a given amount of time, Jamie drove twice as far as Rhonda. Altogether they drove 90 miles. Find the number of miles driven by each.
Step 1: What are we trying to find? The number of miles driven by Jamie and by Rhonda. Step 2: Assign a variable. Since we are looking for two numbers here, we need to choose which one we will assign a variable to. The number of miles driven by either Jamie or Rhonda will work. We need to just choose one and move to Step 3. Let’s assign a variable to represent the number of miles driving by Rhonda
Let’s call it R. Step 3: Write down what the variable represents. Let R = the number of miles driven by Rhonda Step 4: Write an equation. We know that Jamie drove twice as far a Rhonda. As with Example 4, let’s think about this in terms of numbers before jumping into an equation. If Rhonda drives 10 miles, then Jamie will drive twice as far which would be 20. So whatever amount Rhonda drives, Jamie’s amount will be two times that number. We have already decided that the number of miles driven by Rhonda is R, so the number of miles driven by Jamie is 2R. Together they drove a total of 90 miles. So we have (Rhonda) + (Jamie) = 90, or Step 5: Solve the equation. Step 6: Answer the question in the problem The problem asks us to find out how far Rhonda and Jamie drove. The solution to the equation tells us R = 30, which means Rhonda drove 30 miles. Now we have to find out how far Jamie drove. She drove twice as far as Rhonda, so the distance would be 20 miles. Step 7: Check the answer. The answer makes sense and checks in our equation from Step 4.
 Karen works for $6 an hour. A total of 25% of her salary is deducted for taxes and insurance. She is trying to save $450 for a new car stereo and speakers. How many hours must she work to take home $450 if she saves all of her earnings?
Step 1: What are we trying to find? The number of hours Karen needs to work. Step 2: Assign a variable for the number of hours. Let’s call it h. Step 3: Write down what the variable represents. Let h = the number of hours Karen needs to work Step 4: Write an equation. However many hours Karen works, we multiply that number by 6 to find out how much she earns. For example, if she worked, 10 hours, she would make $60 before taxes and insurance. So her salary before taxes and insurance will be 6h. From that amount, we have to subtract the amount taken out for taxes and insurance. 25% of her salary is taken away. We need to write 25% as a decimal which gives 0.25. But we have to take 25% OF her salary or 25% of 6h. Karen’s goal is $450. We can now write an equation.
(Salary)  25%(Salary) = 450
You may wonder why we did not use a dollar sign in the equation. Some students find the extra symbols distracting. It will be necessary to include dollars as part of any answer we may give involving money in this problem. Step 5: Solve the equation. Step 6: Answer the question in the problem The problem asks us to find how many hours Karen needs to work. We decided that h would be the number, so we have h = 100. Karen needs to work 100 hours to reach her goal of $450. Step 7: Check the answer. The answer makes sense and checks in our equation from Step 4.
 The length of a rectangular map is 15 inches and the perimeter is 50 inches. Find the width.
Step 1: What are we trying to find? The width of a rectangle. Step 2: Assign a variable for the width. Let’s call it w. Step 3: Write down what the variable represents. Let w = the width of a rectangle Step 4: Write an equation. We know the length is 15 inches. We also know the perimeter is 50 inches. Perimeter is the distance all the way around a figure. So to go all the way around a rectangle, you have
Perimeter = width + length + width + length.
Since length is 15 inches, width is w, and perimeter is 50, we get
Step 5: Solve the equation. Step 6: Answer the question in the problem. The problem asks us to find the width of a rectangle. We decided that w would represent width, so we have w = 10. The width of the rectangle is 10 inches. Don’t forget your units. Step 7: Check the answer. The answer makes sense and checks in our equation from Step 4.
inches
 The circumference of a circular clock face is 13.12 centimeters more than three times the radius. Find the radius of the face.
Step 1: What are we trying to find? The radius of the face of a circular clock. Step 2: Assign a variable for the radius. Let’s call it r. Step 3: Write down what the variable represents. Let r = the radius of the clock face Step 4: Write an equation. First we need to know a formula that will relate circumference and radius since those are two pieces of information in the problem. The formula for the circumference is . We are told that the circumference is 13.12 centimeters more than three times the radius. Three times the radius translates into 3r. Now we need to add 13.12 to that to get an expression for circumference.
We now have two expressions for circumference. Since the circumference of a circle doesn’t change, these two expressions must be equal. Now we can set up the equation
Step 5: Solve the equation. In some classes your teacher may want you to leave in its exact form rather than approximating the value as 3.14. We will use the approximation here. If your teacher wants you to leave as part of your answer, you should ask how to do that.
Step 6: Answer the question in the problem The problem asks us to find the radius of the clock face. We decided that r would be the radius, so we have r = 4. The radius of the clock face is 4 centimeters. Don’t forget your units. Step 7: Check the answer. The answer makes sense and checks in our equation from Step 4.


