Problem 52 - College Algebra

Question

For the following exercises, 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 \(A E \neq B D .\) $$ \begin{array}{l}{x+A y=1} \\ {x+B y=1}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Set up the system of equations

The given system is: 1. \( x + Ay = 1 \) 2. \( x + By = 1 \) Our task is to solve this system for \(x\) and \(y\) in terms of the constants \(A, B, C, D, E,\) and \(F\).

Step 2: Eliminate x

Subtract the second equation from the first equation to eliminate \(x\): \((x + Ay) - (x + By) = 0\). This simplifies to: \((A - B)y = 0\).

Step 3: Solve for y

Since \((A - B)y = 0\) and given that \(A eq B\), we find that \(y = 0\). This is because the only way for the product of a nonzero term with \(y\) to equal zero is for \(y\) itself to be zero.

Step 4: Substitute y back into one equation

Substitute \(y = 0\) back into the first equation to find \(x\): \(x + A(0) = 1\) which simplifies to \(x = 1\).

Step 5: Verify the solution

Substitute \(x = 1\) and \(y = 0\) back into the second equation to verify the solution: \(1 + B(0) = 1\), which holds true, confirming that our solution \(x = 1\) and \(y = 0\) is correct.

Key Concepts

Solving Linear Equations

Understanding how to solve linear equations can greatly simplify the process of finding unknown variables. Linear equations are mathematical statements of equality involving one or more variables. These equations are called "linear" because they graph as straight lines. When dealing with such equations, our primary aim is to find the values of the unknowns that make these equations true.

A typical system of linear equations might look like this: \ \(x + Ay = 1\) and \ \(x + By = 1\). The challenge is to find values for \(x\) and \(y\) that satisfy both equations simultaneously.

• Start by organizing the given equations.
• Look for strategies, such as elimination or substitution, to simplify and solve these equations.
• Focus on reducing the number of variables in each step until only one remains, making it easier to solve directly.

Variable Elimination

One effective approach when solving systems of equations is variable elimination. This method helps us to cancel out one variable, making it possible to solve for the other.

In our example system \ \(x + Ay = 1\) and \ \(x + By = 1\), we eliminate \(x\) by subtracting one equation from the other. When we subtract, the terms involving \(x\) cancel out, since \(x + Ay - (x + By) = (A-B)y = 0\).

• We assume \(A eq B\), allowing \((A - B)\) to be non-zero.
• With \((A - B)y = 0\), \(y\) must be zero.

Now, \(x\) can be found by substituting \(y = 0\) back into any original equation, simplifying the solution process.

Verification of Solutions

After finding a solution to a system of equations, it is crucial to verify that the solution satisfies all given equations. Verification ensures that the solution is correct and consistent for all parts of the system.

In the solution \ \(x = 1\) and \ \(y = 0\), we substitute these values back into the original equations. For example, substituting into \ \(x + By = 1\) results in \ \(1 + B(0) = 1\), verifying that our solution is correct.

• Always check each equation in the system.
• Ensure that both \(x\) and \(y\) satisfy the individual equations when substituted back.
• This step is essential to confirm the accuracy of the entire solving process.

By practicing this verification process, you build confidence in your solutions and ensure that no mistakes were made during the calculations.