Problem 53 - 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 \(A E \neq B D .\) $$ \begin{array}{l}{A x+y=0} \\ {B x+y=1}\end{array} $$

Step-by-Step Solution

Answer

\(x = \frac{1}{B - A}\), \(y = -\frac{A}{B - A}\)

Step 1: Set Up the Equations

We start with the given system of equations: 1. \( A x + y = 0 \) 2. \( B x + y = 1 \)

Step 2: Eliminate y by Subtraction

Subtract the first equation from the second equation to eliminate \( y \). \[ (B x + y) - (A x + y) = 1 - 0 \] This simplifies to: \[ (B - A)x = 1 \]

Step 3: Solve for x

Isolate \( x \) by dividing both sides of the equation \( (B - A)x = 1 \) by \( B - A \). \[ x = \frac{1}{B - A} \]

Step 4: Substitute x into the First Equation

Substitute \( x = \frac{1}{B - A} \) back into the first equation \( A x + y = 0 \). \[ A\left(\frac{1}{B - A}\right) + y = 0 \] Simplify to find \( y \): \[ \frac{A}{B - A} + y = 0 \] \[ y = -\frac{A}{B - A} \]

Step 5: Solution of the System

The solution to the system is \[ x = \frac{1}{B - A}, \quad y = -\frac{A}{B - A} \]

Key Concepts

Solving Linear Equations

Solving linear equations involves finding the values of variables that satisfy all equations in a system. Linear equations are equations of the first order. This means they will graph as straight lines on a coordinate plane. In our example, we are given a system of linear equations:

• \( A x + y = 0 \)
• \( B x + y = 1 \)

These equations represent two lines. Our task is to find the intersection of these two lines, representing the solution. When solving systems of linear equations, you can use several methods, including substitution and elimination, which will help us find solutions more effectively.

Substitution Method

The substitution method is a technique for solving systems of equations by solving one equation for one variable and then substituting that solution into the other equation. This method simplifies the equations step by step.In the given system:

• We first solve one of the equations for \( y \) or \( x \).

Typically, if one of the equations is already easy to solve for one variable, that equation gets used.However, for our solution, the elimination method makes more sense due to its straightforward approach to eliminating variables efficiently.

Elimination Method

The elimination method involves combining equations in a system to cancel out one of the variables. It is a powerful technique for solving systems of equations, especially suitable when coefficients are easy to manipulate.For the given system:

• We subtract Equation 1 from Equation 2: \( (B x + y) - (A x + y) = 1 - 0 \).
• This cancels out \( y \) and simplifies to \((B - A)x = 1\).

Solving for \( x \) gives us \( x = \frac{1}{B - A} \). We then substitute this value back into one of the original equations to find \( y \). This method is often preferred for its systematic approach, which helps validate solutions efficiently.

Linear Algebra Concepts

Linear algebra provides the foundational framework for working with linear equations and systems. Core concepts include vectors, matrices, and operations such as addition and multiplication.In terms of solving sets of equations:

• You can represent equations using matrices. This representation facilitates operations such as row reduction, which are central in more advanced solutions beyond substitution and elimination.
• The conditions \( A eq B \) and \( A E eq B D \) are crucial. They ensure that the equations are independent and thus solvable, as dependent or equivalent equations might lead to infinite solutions or no solutions at all.

Understanding these concepts helps in interpreting results and checking whether systems have unique solutions, infinite solutions, or none.