Problem 53 - 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 \(A E \neq B D\). \(A x+y=0\) \(B x+y=1\)

Step-by-Step Solution

Answer

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

Step 1: Express y in terms of x from the first equation

The first equation is \( Ax + y = 0 \). We can solve for \( y \) by isolating it on one side of the equation. This gives us \( y = -Ax \).

Step 2: Substitute y into the second equation

Now we substitute \( y = -Ax \) into the second equation \( Bx + y = 1 \). This results in \( Bx - Ax = 1 \).

Step 3: Simplify and solve the equation for x

The equation from Step 2 is \( Bx - Ax = 1 \). Factor \( x \) out of the terms on the left side to get \( (B - A)x = 1 \). Since \( A eq B \), divide both sides by \( B - A \) to solve for \( x \), thus \( x = \frac{1}{B-A} \).

Step 4: Substitute x back to find y

Now that we have \( x = \frac{1}{B-A} \), substitute this into the expression for \( y \) from Step 1. Thus, \( y = -A\left(\frac{1}{B-A}\right) = \frac{-A}{B-A} \).

Step 5: Write the solution in terms of A, B, C, D, E, F

Based on the calculations, the solution to the system of equations \(Ax + y = 0\) and \(Bx + y = 1\) is \(x = \frac{1}{B-A}\) and \(y = \frac{-A}{B-A}\).

Key Concepts

Substitution Method

The substitution method is a powerful technique used for solving linear systems of equations. It involves solving one equation for one variable, and then substituting that expression into another equation.

• In our problem, the original system has two equations: \( Ax + y = 0 \) and \( Bx + y = 1 \).
• First, we isolated \( y \) from the first equation, giving us \( y = -Ax \).
• This expression for \( y \) is then substituted into the second equation.

By substituting \( y = -Ax \) into \( Bx + y = 1 \), we effectively reduce the system into a single equation in terms of \( x \). This simplifies the process of solving for the variables. The substitution method is especially useful when the system easily allows one of the variables to be isolated.

Solving for Variables

Solving for variables involves simplifying and rearranging equations to find the value of unknowns. The goal is to get the variable on one side of the equation.

• After substituting \( y = -Ax \) into \( Bx + y = 1 \), we reconfirm the equation as \( Bx - Ax = 1 \).
• To solve for \( x \), notice that \( x \) can be factored out on the left-hand side: \((B - A)x = 1 \).

Because the problem states that \( A eq B \), dividing both sides of the equation by \( B - A \) gives us \( x = \frac{1}{B-A} \). This process highlights key algebra principles like factoring and dividing to isolate a variable. Understanding these steps is crucial for effectively unraveling more complex systems.

Algebraic Manipulation

Algebraic manipulation is the technique of using algebraic operations to transform an equation into a desired form. It plays a critical role in solutions like our current problem.

• Starting with \( Ax + y = 0 \), algebraic manipulation helped us rearrange it to find \( y = -Ax \).
• In the transformed equation \( (B-A)x = 1 \), we used algebraic manipulation to factor \( x \).

Finally, to solve for \( y \), we substitute \( x = \frac{1}{B-A} \) back into \( y = -Ax \). This gives us \( y = \frac{-A}{B-A} \). Understanding algebraic manipulation is essential to solve equations efficiently, as it involves accurately applying operations to reach a simpler or solved form.