Word Problems With Variables On Both Sides

Navigating the complexities of algebra often leads us to the realm of word problems. These problems, seemingly daunting at first, are opportunities to translate real-world scenarios into mathematical equations. Among the various types of algebraic word problems, those involving variables on both sides of the equation present a unique and valuable challenge. Mastering this skill unlocks a deeper understanding of algebraic manipulation and its practical applications.

Understanding the Basics: What are Variables on Both Sides?

Before diving into problem-solving, let's clarify what we mean by "variables on both sides." In an algebraic equation, a variable is a symbol (usually a letter like x, y, or z) that represents an unknown quantity. When the same variable appears on both sides of the equals sign (=), it signifies that the unknown quantity influences both expressions in the equation.

For example, consider the equation: 3x + 5 = x - 1. Here, the variable x appears on both the left side (3x + 5) and the right side (x - 1) of the equation. Solving such equations requires a strategic approach to isolate the variable on one side and determine its value.

The Core Strategy: Isolating the Variable

The primary goal when solving word problems with variables on both sides is to isolate the variable on one side of the equation. This involves a series of algebraic manipulations using the properties of equality. The key principles to remember are:

• Addition/Subtraction Property of Equality: You can add or subtract the same value from both sides of the equation without changing its balance.
• Multiplication/Division Property of Equality: You can multiply or divide both sides of the equation by the same non-zero value without changing its balance.
• Combining Like Terms: Simplify each side of the equation by combining terms that contain the same variable or are constants.

Step-by-Step Guide to Solving Word Problems with Variables on Both Sides

Let's break down the process of solving word problems with variables on both sides into manageable steps:

1. Read and Understand the Problem: Carefully read the problem statement and identify the unknown quantities you need to find. Determine the relationship between these quantities based on the information provided in the problem. Look for keywords or phrases that suggest mathematical operations (e.g., "sum," "difference," "product," "quotient," "is," "equals").

2. Define the Variable(s): Assign a variable to represent the unknown quantity you are trying to find. If there are multiple unknowns, try to express them in terms of a single variable whenever possible. This will simplify the equation.

3. Translate the Words into an Equation: This is the crucial step where you convert the word problem into a mathematical equation. Use the information you gathered in step 1 and the variable(s) you defined in step 2 to write an equation that accurately represents the problem. Pay close attention to the order of operations and the relationships between the quantities.

4. Simplify the Equation: Before isolating the variable, simplify both sides of the equation by combining like terms. This will make the equation easier to manipulate and reduce the chances of making errors.

5. Isolate the Variable: Use the addition/subtraction property of equality to move all terms containing the variable to one side of the equation and all constant terms to the other side. Remember to perform the same operation on both sides to maintain the balance.

6. Solve for the Variable: Once the variable is isolated, use the multiplication/division property of equality to solve for its value. Divide both sides of the equation by the coefficient of the variable.

7. Check Your Solution: Substitute the value you found for the variable back into the original equation to verify that it satisfies the equation. Also, make sure your answer makes sense in the context of the original word problem. If your solution doesn't check out, re-examine your work to identify any errors.

8. Answer the Question: Finally, answer the question posed in the word problem using the value you found for the variable. Make sure your answer is clear, concise, and includes the appropriate units (if applicable).

Example Problems with Detailed Solutions

Let's illustrate the problem-solving process with a few example problems:

Problem 1:

John and Mary are saving money. John has already saved $50 and plans to save $10 per week. Mary has saved $20 and plans to save $15 per week. After how many weeks will John and Mary have saved the same amount of money?

Solution:

1. Understand the Problem: We need to find the number of weeks after which John and Mary will have the same amount of savings.

2. Define the Variable: Let w represent the number of weeks.

3. Translate into an Equation:

   • John's savings: 50 + 10w
   • Mary's savings: 20 + 15w
   • Equation: 50 + 10w = 20 + 15w

4. Simplify the Equation: The equation is already simplified.

5. Isolate the Variable:

   • Subtract 10w from both sides: 50 = 20 + 5w
   • Subtract 20 from both sides: 30 = 5w

6. Solve for the Variable:

   • Divide both sides by 5: w = 6

7. Check the Solution:

   • John's savings after 6 weeks: 50 + 10(6) = 50 + 60 = $110
   • Mary's savings after 6 weeks: 20 + 15(6) = 20 + 90 = $110

8. Answer the Question: After 6 weeks, John and Mary will have saved the same amount of money ($110).

Problem 2:

Two companies, Alpha and Beta, offer different salary packages. Alpha offers a base salary of $40,000 plus a commission of $500 for each sale. Beta offers a base salary of $30,000 plus a commission of $800 for each sale. How many sales must an employee make for the two salary packages to be equal?

Solution:

1. Understand the Problem: We need to find the number of sales that will result in the same total salary for both companies.

2. Define the Variable: Let s represent the number of sales.

3. Translate into an Equation:

   • Alpha's salary: 40000 + 500s
   • Beta's salary: 30000 + 800s
   • Equation: 40000 + 500s = 30000 + 800s

4. Simplify the Equation: The equation is already simplified.

5. Isolate the Variable:

   • Subtract 500s from both sides: 40000 = 30000 + 300s
   • Subtract 30000 from both sides: 10000 = 300s

6. Solve for the Variable:

   • Divide both sides by 300: s = 33.33 (approximately)

7. Check the Solution: Since we can't have a fraction of a sale, we need to consider 33 and 34 sales.

   • Alpha's salary with 33 sales: 40000 + 500(33) = $56,500
   • Beta's salary with 33 sales: 30000 + 800(33) = $56,400
   • Alpha's salary with 34 sales: 40000 + 500(34) = $57,000
   • Beta's salary with 34 sales: 30000 + 800(34) = $57,200

   Since the salaries are closest with 33 sales, and the question asks for the number of sales where the packages are equal, we should round up to 34. However, technically they are never perfectly equal. The problem should be reworded to ask "how many sales must an employee make for Beta's salary to be greater than Alpha's salary?" The answer would then be 34. For the purposes of illustrating the variable isolation process, we'll stick with the imperfect equality.

8. Answer the Question: An employee must make approximately 33.33 sales for the two salary packages to be equal. (Note: In a real-world scenario, since sales are discrete, the salaries will never be exactly equal). Because of this imperfection, we will round to 34 sales.

Problem 3:

A rectangle has a length that is 5 cm longer than its width. Another rectangle has a length that is twice its width. If both rectangles have the same perimeter, and the second rectangle has a width of 6 cm, what is the width of the first rectangle?

Solution:

1. Understand the Problem: We are given information about two rectangles and their perimeters. We need to find the width of the first rectangle.

2. Define the Variable: Let w represent the width of the first rectangle. The length of the first rectangle is then w + 5.

3. Translate into an Equation:

   • Perimeter of the first rectangle: 2w + 2(w + 5) = 2w + 2w + 10 = 4w + 10
   • Width of the second rectangle: 6 cm
   • Length of the second rectangle: 2 * 6 = 12 cm
   • Perimeter of the second rectangle: 2(6) + 2(12) = 12 + 24 = 36 cm
   • Equation: 4w + 10 = 36

4. Simplify the Equation: The equation is already simplified.

5. Isolate the Variable:

   • Subtract 10 from both sides: 4w = 26

6. Solve for the Variable:

   • Divide both sides by 4: w = 6.5

7. Check the Solution:

   • Width of the first rectangle: 6.5 cm
   • Length of the first rectangle: 6.5 + 5 = 11.5 cm
   • Perimeter of the first rectangle: 2(6.5) + 2(11.5) = 13 + 23 = 36 cm
   • Perimeter of the second rectangle: 36 cm (as calculated previously)

8. Answer the Question: The width of the first rectangle is 6.5 cm.

Advanced Strategies and Considerations

While the step-by-step guide provides a solid foundation, here are some advanced strategies and considerations for tackling more complex word problems:

• Look for Hidden Relationships: Some problems may not explicitly state the relationship between the variables. You might need to infer the relationship based on the context of the problem or use your knowledge of geometry, physics, or other subjects.

• Using Systems of Equations: If a problem involves multiple unknowns and multiple relationships between them, you might need to use a system of equations to solve it. This involves setting up two or more equations and solving them simultaneously to find the values of all the variables.

• Dealing with Inequalities: Some word problems may involve inequalities instead of equalities. In these cases, you'll need to use the properties of inequalities to solve for the variable. Remember that multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality sign.

• Working with Rate, Time, and Distance: Many word problems involve rate, time, and distance. Remember the fundamental relationship: Distance = Rate × Time. Use this formula to set up equations and solve for the unknown quantities.

• Age Problems: Age problems often involve comparing the ages of people at different points in time. Set up equations that represent the ages of the people at different times and use the information given in the problem to solve for the unknown ages.

• Mixture Problems: Mixture problems involve combining two or more substances with different properties (e.g., concentration, price) to create a mixture with a desired property. Set up equations that represent the amounts of each substance and their properties and use the information given in the problem to solve for the unknown quantities.

Common Mistakes to Avoid

Solving word problems with variables on both sides requires careful attention to detail. Here are some common mistakes to avoid:

• Misinterpreting the Problem: Failing to fully understand the problem statement can lead to incorrect equations and solutions. Read the problem carefully and identify the key information and relationships.

• Incorrectly Defining Variables: Choosing the wrong variable or failing to define it clearly can make it difficult to translate the problem into an equation. Make sure your variables accurately represent the unknown quantities you are trying to find.

• Making Errors in Algebraic Manipulation: Careless mistakes in algebraic manipulation, such as forgetting to distribute a negative sign or combining unlike terms, can lead to incorrect solutions. Double-check your work carefully.

• Forgetting to Check the Solution: Failing to check your solution can result in accepting an incorrect answer. Always substitute your solution back into the original equation to verify that it satisfies the equation and makes sense in the context of the problem.

• Not Answering the Question: Finding the value of the variable is not the final step. Make sure you answer the question posed in the word problem using the value you found.

Practice Problems

To solidify your understanding of word problems with variables on both sides, try solving the following practice problems:

1. Sarah and Tom are running a race. Sarah starts 10 meters ahead of Tom. Sarah runs at a speed of 5 meters per second, and Tom runs at a speed of 7 meters per second. After how many seconds will Tom catch up to Sarah?

2. A store is having a sale on shirts. All shirts are marked down by 20%. John has a coupon for an additional 10% off the sale price. Mary has a coupon for $5 off the original price. If both John and Mary buy the same shirt, and the original price of the shirt is p, for what value of p will John and Mary pay the same amount?

3. Two cars leave the same city at the same time and travel in opposite directions. One car travels at a speed of 60 miles per hour, and the other car travels at a speed of 70 miles per hour. After how many hours will the two cars be 390 miles apart?

Conclusion

Word problems with variables on both sides are a fundamental concept in algebra that bridges the gap between abstract mathematical concepts and real-world applications. By mastering the step-by-step problem-solving process, understanding advanced strategies, and avoiding common mistakes, you can confidently tackle these problems and unlock a deeper understanding of algebraic manipulation. Remember that practice is key to success. The more you practice solving word problems, the more comfortable and confident you will become. So, embrace the challenge, hone your skills, and enjoy the journey of mastering algebra!