Problem 51 - College Algebra

Question

Solve each system in terms of \(A, B, C, D, E,\) and \(F\) where \(A-F\) are nonzero numbers. Note that \(A \neq B\) and \(A E \neq B D\). $$ \begin{array}{l} x+y=A \\ x-y=B \end{array} $$

Step-by-Step Solution

Answer

\( x = \frac{A + B}{2}, y = \frac{A - B}{2} \)

Step 1: Solve for x

Let's solve the system of equations for \( x \). First, add the two equations together: \[ (x + y) + (x - y) = A + B. \]The \( y \) terms will cancel out, so we have:\[ 2x = A + B. \]Divide both sides by 2 to find \( x \):\[ x = \frac{A + B}{2}. \]

Step 2: Solve for y

Now, solve for \( y \). Subtract the second equation from the first:\[ (x + y) - (x - y) = A - B. \]This cancels out the \( x \) terms, leaving:\[ 2y = A - B. \]Divide both sides by 2 to get \( y \):\[ y = \frac{A - B}{2}. \]

Key Concepts

Linear Equations

Linear equations are the simplest form of algebraic equations and are foundational in understanding more complex topics. A linear equation is an equation of the first degree, which typically involves two variables and a constant. In a two-variable context, like our exercise, each equation forms a straight line when graphed on a coordinate plane.
For instance, in the system of equations given:

• Equation 1: \( x + y = A \)
• Equation 2: \( x - y = B \)

Each of these equations is a line in a two-dimensional space.
Linear equations have the property that their graph is a straight line, indicating that there are no squared (or higher order) terms. Their solutions are the points where these lines intersect. Understanding these basic ideas about linear equations is crucial in solving them.

Solving Equations

When we talk about solving equations, we are looking for values of variables that satisfy the equations given. In a system of equations, you typically seek a set of values that fulfills all the equations simultaneously.
For the steps provided in the solution:

• The first step was to find \( x \) by adding the two equations: \( (x + y) + (x - y) = A + B \). The goal is to simplify and eliminate one variable.
• The second step required finding \( y \) by subtracting the second equation from the first: \( (x + y) - (x - y) = A - B \).

Each equation manipulation involved either adding or subtracting to eliminate a variable, reducing the equation to one unknown, and then making that unknown the subject to solve for it.
This methodical approach is a reliable way to tackle systems of linear equations.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate variables and solve them. This skill is essential for solving systems of equations as it allows you to move terms around systematically. Here's how it worked in the solution:

• Addition and subtraction of equations were used to remove individual terms and simplify the systems.
• Division by a coefficient was the final step to isolate the variable of interest, in this case, dividing by 2 to solve for \( x \) and \( y \).

To manipulate algebraic expressions effectively, remember these steps:

• First identify which variable you can eliminate through addition or subtraction.
• Simplify the resulting equation as much as possible.
• Rearrange the terms to make a single variable the subject of the equation.

With practice, algebraic manipulation becomes intuitive, providing a powerful toolkit for addressing various mathematical problems.