Two workers were hired to begin work at the same time. Worker A's contract called for a starting salary of \(\$ 20,000\) with an increase of \(\$ 800\) after each year of employment. Worker \(B\) 's contract called for a starting salary of \(\$ 15,200\) with an increase of \(\$ 2,000\) after each year of employment (that is, the number of full increases each worker has received), which of the number of years until \(B\) 's yearly salary equals A's yearly salary? A. \(20,000+800 x=15,200+2,000 x\) B. \(20,000+2,000 x=15,200+800 x\) C. \((20,000+800) x=(15,200+2,000) x\) D. \((2,000+800) x=20,000-15,200\) E. \((2,000-800) x=20,000+15,200\)

Linear equations are foundational in algebra and represent relationships where there is a constant rate of change. They can be used to model real-life situations, like the one we see with the worker salary problem. A linear equation looks like this: \( ax + b = c \), where \( a \), \( b \), and \( c \), are constants, and \( x \) is the variable we aim to solve for.

These equations can graphically be represented as straight lines when plotted on a Cartesian plane, hence the term 'linear.' The key to understanding and solving these equations lies in isolating the variable \( x \) on one side of the equation. This often involves operations such as adding, subtracting, multiplying, or dividing entire equations by numbers, which helps to simplify and ultimately solve the problem at hand.

In our exercise, the linear equations describe the salary progression over time for each worker. By understanding these principles, we can see how the increments add up yearly, leading to linear growth in salaries.

Solving for \( x \) is a central skill in algebra. It means finding the value of the variable that makes the equation true. Most commonly, we're dealing with numerical equations, but algebra allows us to solve for \( x \) even when we have an equation made up of several different variables or constants.

In the context of our problem, solving for \( x \) involves manipulations to both sides of the equation so that \( x \) is by itself on one side, and all the constants are on the other. This often entails subtracting or adding terms from both sides to get rid of any extra \( x \) terms and then dividing or multiplying to isolate \( x \) completely.

Example Steps for Solving for \( x \) in our Exercise

• Equalize equations for both workers' salaries.
• Subtract constants and like terms on both sides to gather \( x \) terms on one side.
• Combine like terms to simplify the equation.
• Finally, divide by the coefficient of \( x \) to find the value of \( x \) that balances the equation.

Following these steps leads us to the solution where \( x \) equals the number of years until the salaries of two workers are equal.

Algebraic expressions are a mix of numbers, variables (like \( x \) in our problem), and operation symbols that show the relationship between these elements. They are the building blocks of equations and complex formulas. An algebraic expression might look complicated at first glance, but they're just a way to represent problems or situations mathematically.

In our salary problem, \( A_{salary} = 20,000 + 800x \) and \( B_{salary} = 15,200 + 2,000x \) are both algebraic expressions. They illustrate the salary calculation per year for each worker. The constant term in these expressions (\20,000\ and \(15,200\respectively) represents the starting salaries, while the coefficients of \( x \), \(800\and \(2,000\represent the yearly raise for each worker.

Grasping the structure of these expressions is important as it allows us to understand the problem thoroughly and apply the correct methods to find solutions to the questions posed. Mastering algebraic expressions lays the groundwork for tackling more advanced mathematics.