Problem 52 - College Algebra with Corequisite Support

Question

For the following exercises, solve each system in terms of \(A, B, C, D, E, \quad\) and \(F\) where \(A-F\) are nonzero numbers. Note that \(A \neq B\) and \(A E \neq B D\). \(x+A y=1\) \(x+B y=1\)

Step-by-Step Solution

Answer

The solution is \(x = 1\) and \(y = 0\).

Step 1: Set Up the System of Equations

We are given the system of equations: 1. \( x + Ay = 1 \) 2. \( x + By = 1 \)

Step 2: Eliminate Variable x

Subtract the second equation from the first: \[ (x + Ay) - (x + By) = 1 - 1 \] This simplifies to: \[ Ay - By = 0 \]

Step 3: Factor and Solve for y

Factor out \(y\) from the equation: \[ y(A - B) = 0 \] Since \(A eq B\), \(A-B eq 0\), hence \(y = 0\).

Step 4: Solve for x

Substitute \(y = 0\) back into one of the original equations, for example the first one: \[ x + A(0) = 1 \] This simplifies to \(x = 1\).

Step 5: Verify the Solution

Check the solution in the second equation: \[ x + B(0) = 1 \] This simplifies to \(x = 1\), verifying \(x = 1\) and \(y = 0\) as the solution.

Key Concepts

Elimination Method

The elimination method is a powerful tool when it comes to solving systems of equations. Its key idea revolves around eliminating one of the variables in order to simplify the system and make it easier to solve for the remaining variables. In our example, we want to eliminate the variable \(x\). Here's the process:

• Align the equations so that the corresponding terms are in the same order.
• Apply operations such as addition, subtraction, multiplication, or division to eliminate a variable. In this case, we subtract one equation from the other.

This subtraction eliminates the \(x\) term from both equations:

• Equation 1: \(x + Ay = 1\)
• Equation 2: \(x + By = 1\)
• Subtract Equation 2 from Equation 1 to get: \((x + Ay) - (x + By) = 1 - 1\)
• Simplified result is: \(Ay - By = 0\)

By eliminating \(x\), we now have a single equation with one variable, \(y\). This drastically simplifies the problem and paves the way to finding the solution.

Solving Equations

Solving equations is the act of finding the value(s) of the variable(s) that satisfy the given equation(s). Once we isolate a variable using the elimination method, the next step is to solve for it. In our example, after eliminating \(x\), we are left with:

• Equation: \(Ay - By = 0\)
• This leads to \((A - B)y = 0\)

To solve this equation for \(y\), notice that it's a product equaling zero. For this product to hold true given \(A eq B\), we conclude:

• \(y = 0\)

Substitute back to find \(x\) using any of the original equations:

• For \(x + Ay = 1\) and \(y = 0\):
• \(x + A(0) = 1\)
• Thus, \(x = 1\)

Verification is an essential step. We check our solutions by substituting \(y = 0\) and \(x = 1\) back into the second equation \((x + By = 1)\) and confirming they satisfy it.

Algebraic Expressions

Understanding algebraic expressions is crucial when working with systems of equations. In any algebraic manipulation, expressions are the foundation. They consist of constants, variables, and operations such as addition and subtraction.
In our exercise, we deal with algebraic expressions like \(x + Ay\) and \(x + By\). Here:

• "x" and "y" are variables, while "A" and "B" are constants.
• The expressions are linear, meaning each variable is to the power of one.
• This simplicity allows us to solve the system using straightforward techniques such as elimination.

Factoring is an important operation in algebra. When we reached the equation \(Ay - By = 0\), factoring out \(y\) resulted in \(y(A - B)\) which simplified the equation and allowed us to determine \(y = 0\).
Understanding these expressions both in structure and application is essential to effectively solving equations. Whether it's recognizing how to rearrange terms or factor constants and variables, mastering expressions is key to success in algebra.