Word Problems Involving Linear Equations

Introduction

Key Concepts

Understanding Linear Equations

A linear equation in one variable is an algebraic expression where the variable has a power of one. It can be expressed in the form: $$ax + b = c$$ where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. The solution to this equation is the value of \( x \) that satisfies the equality.

Formulating Word Problems into Linear Equations

The first step in solving word problems involving linear equations is translating the given information into a mathematical equation. This involves identifying the variables, constants, and the relationship between them. Consider the following example:

*Example:* Jane has twice as many apples as Tom. Together, they have 18 apples. How many apples does each person have?

Let \( x \) represent the number of apples Tom has. Therefore, Jane has \( 2x \) apples. The equation can be formulated as: $$x + 2x = 18$$ Solving for \( x \): $$3x = 18$$ $$x = 6$$ Thus, Tom has 6 apples, and Jane has \( 2 \times 6 = 12 \) apples.

Solving Linear Equations

Once the equation is formulated, solving for the variable involves isolating it on one side of the equation. The basic steps include:

1. Combine like terms on both sides.
2. Use inverse operations to isolate the variable.
3. Simplify to find the value of the variable.

*Example:* A car rental company charges a flat fee of $50 plus $20 per day. If a customer is charged $90, how many days did they rent the car?

Let \( d \) represent the number of days. The equation is: $$50 + 20d = 90$$ Subtract 50 from both sides: $$20d = 40$$ Divide by 20: $$d = 2$$ Therefore, the customer rented the car for 2 days.

Applications of Linear Equations in Real Life

Linear equations are prevalent in various real-life scenarios, including financial calculations, measurement conversions, and planning. Understanding how to model these situations mathematically allows for efficient problem-solving and decision-making.

*Example:* If a smartphone costs $300 and a tablet costs $450, how many smartphones and tablets can be purchased with $2400 if smartphones are bought?

Let \( s \) represent the number of smartphones, and \( t \) the number of tablets. The equation is: $$300s + 450t = 2400$$ This can be simplified to: $$2s + 3t = 16$$ By choosing values for \( s \) and solving for \( t \), feasible solutions can be found.

Graphical Representation

Graphing linear equations provides a visual understanding of the relationship between variables. The slope-intercept form \( y = mx + b \) depicts a straight line where \( m \) is the slope and \( b \) is the y-intercept. In word problems, this can represent rates of change and fixed costs.

*Example:* Represent the equation \( y = 20x + 50 \) on a graph, where \( y \) is the total cost and \( x \) is the number of days rented.

The y-intercept is 50, indicating the fixed fee, and the slope is 20, representing the daily rate. Plotting this equation helps visualize how the total cost increases with each additional day.

System of Linear Equations

Some word problems require solving systems of linear equations, where multiple equations are used to find the values of multiple variables simultaneously.

*Example:* A school is selling tickets for a play. If adult tickets cost $10 and student tickets cost $6, and the total revenue is $160 from selling 20 tickets, how many of each ticket were sold?

Let \( a \) be the number of adult tickets and \( s \) the number of student tickets. The system of equations is: $$a + s = 20$$ $$10a + 6s = 160$$ Solving the system gives: $$a = 10$$ $$s = 10$$ Thus, 10 adult tickets and 10 student tickets were sold.

Common Challenges and Strategies

Students often face challenges in setting up the correct equations and interpreting the results within the context of the problem. To overcome these, the following strategies are recommended:

• Careful Reading: Understand the problem thoroughly before setting up equations.
• Identifying Variables: Clearly define what each variable represents.
• Simplification: Simplify the equations to make them easier to solve.
• Checking Solutions: Substitute the solution back into the original equation to verify its correctness.

Practicing a variety of word problems enhances proficiency and confidence in solving linear equations.

Example Problems

*Problem 1:* A farmer has 150 meters of fencing and wants to build a rectangular enclosure. If the length is 10 meters more than the width, what are the dimensions of the enclosure?

Let \( w \) be the width, then the length \( l = w + 10 \). The perimeter equation is: $$2w + 2(w + 10) = 150$$ Simplifying: $$2w + 2w + 20 = 150$$ $$4w = 130$$ $$w = 32.5$$ Thus, the length \( l = 32.5 + 10 = 42.5 \) meters.

*Problem 2:* A cell phone plan costs $25 per month plus $0.10 per text message. If a customer is billed $45 for a month, how many text messages did they send?

Let \( t \) be the number of text messages. The equation is: $$25 + 0.10t = 45$$ Subtract 25: $$0.10t = 20$$ $$t = 200$$ The customer sent 200 text messages.

Comparison Table

Summary and Key Takeaways