Problem 52 - 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 \(A E \neq B D\). $$ \begin{array}{l} x+A y=1 \\ x+B y=1 \end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the System of Equations

We are given the system of two equations: \( x + Ay = 1 \) and \( x + By = 1 \). Our task is to solve for \( x \) and \( y \) in terms of the constants \( A \) and \( B \), while given that these constants are nonzero and \( A eq B \).

Step 2: Subtract the Equations

To eliminate \( x \) from the equations, subtract the second equation from the first one: \[(x + Ay) - (x + By) = 1 - 1\]This simplifies to:\[Ay - By = 0\]

Step 3: Factor and Simplify the Resulting Equation

Factor out \( y \) from the equation obtained:\[(A - B)y = 0\]Since \( A eq B \), the only solution for this equation is:\[y = 0\]

Step 4: Substitute Back to Find x

Substitute \( y = 0 \) back into one of the original equations. Using the first equation \( x + Ay = 1 \):\[x + A(0) = 1\]Simplifying gives:\[x = 1\]

Step 5: Verify the Solution

Substitute \( x = 1 \) and \( y = 0 \) back into the second equation to verify that the solution holds for both equations:\[x + By = 1 \Rightarrow 1 + B(0) = 1\]This checks out, confirming that the solution \( x = 1 \) and \( y = 0 \) satisfies both equations.

Key Concepts

Solving Linear Equations

To solve a system of linear equations means to find the values of variables that satisfy all equations simultaneously. Let's explore the system given in the exercise:

• Equation 1: \( x + Ay = 1 \)
• Equation 2: \( x + By = 1 \)

Our goal is to find the values of \( x \) and \( y \) such that both equations are true. In this problem, the method involves subtraction to eliminate one of the variables. This is a common technique in solving systems of equations.
We start by subtracting the second equation from the first, which effectively removes \( x \) from the equation, simplifying our path to find \( y \).
Once \( y \) is determined, substituting back helps us find \( x \). This two-step process of elimination and substitution is key in solving systems with two variables.

Algebraic Manipulation

Algebraic manipulation plays a crucial role in simplifying and solving equations. It involves rearranging and transforming the equations using operations such as addition, subtraction, multiplication, and division to isolate variables.
In our example, after subtracting the second equation from the first: \[(A - B)y = 0\]We need to manipulate this equation to find \( y \). Since \( A eq B \), it ensures that we do not divide by zero when simplifying \[(A - B)\]. We can safely divide through by \(A - B\) resulting in \( y = 0 \).
This step requires a good grasp on mathematical operations and the properties of equality, ensuring we maintain the balance of the equation while we solve it. Correct algebraic manipulation is essential in the solution process.

Verification of Solutions

Verification of solutions is a critical step in ensuring the solutions obtained are correct and applicable to the original problem.
Once we determine that \( y = 0 \) and substitute back to find \( x = 1 \), we verify by substituting these values back into both original equations:

• For the first equation: \( x + Ay = 1 \Rightarrow 1 + A(0) = 1 \)
• For the second equation: \( x + By = 1 \Rightarrow 1 + B(0) = 1 \)

Both substitutions confirm the truth of the equations, hence verifying the solutions \( x = 1 \) and \( y = 0 \) are indeed correct.
Verification is invaluable as it shows that every step undertaken is not only correct in theory but also practically valid for the initial problem. It ensures no errors occurred during algebraic manipulation and confirms the solution meets the variables' constraints.