Problem 54 - 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 \(A E \neq B D\). \(A x+B y=C\) \(x+y=1\)

Step-by-Step Solution

Answer

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

Step 1: Express y in terms of x

From the equation of the second system, \( x + y = 1 \), express \( y \) in terms of \( x \). We rearrange the equation to get: \( y = 1 - x \).

Step 2: Substitute y in the first equation

Replace \( y \) in the first equation \( A x + B y = C \) with \( 1 - x \) from the previous step. This gives us: \[ A x + B(1 - x) = C \]

Step 3: Distribute and simplify

Expand the equation: \[ A x + B - B x = C \] Combine like terms: \[ (A - B)x + B = C \]

Step 4: Solve for x

Rearrange the equation to solve for \( x \): \[ (A - B)x = C - B \] Thus, \( x = \frac{C - B}{A - B} \).

Step 5: Solve for y using x

Substitute \( x = \frac{C - B}{A - B} \) back into the expression \( y = 1 - x \): \[ y = 1 - \frac{C - B}{A - B} \] \[ y = \frac{A - B}{A - B} - \frac{C - B}{A - B} \] \[ y = \frac{A - C}{A - B} \]

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a cornerstone skill in solving systems of equations. It involves rearranging and simplifying equations to make them easier to solve. In this context, algebraic manipulation allows us to isolate variables and simplify expressions for more straightforward calculations.
First, it's important to understand that every algebraic equation maintains equality. This means you can perform the same operation on both sides of the equation without changing its balance.

• Addition or subtraction can be used to move terms from one side of an equation to the other. For example, moving a term with a variable to one side can help isolate that variable.
• Multiplication or division is typically used to simplify coefficients of variables, especially when trying to isolate a variable.

In our exercise, the equation was rearranged by algebraic manipulation during several steps to express one variable in terms of the other. This rearrangement made it easier to substitute and solve the equations.

Linear Equations

Linear equations are equations of the first order, meaning they have no squares, cubes, or higher powers of the variables they contain. They are expressed in the form: Ax + By = C, where A, B, and C are constants.
These equations, when graphed, form straight lines. The solution to a system of linear equations is the point where the lines intersect. If we have two linear equations with two variables, like in this exercise, we can find their intersection point by solving the equations simultaneously.
Within our example, we have two linear equations:

• First Equation: \( A x + B y = C \)
• Second Equation: \( x + y = 1 \)

Solving these equations involves finding common values of \(x\) and \(y\) that satisfy both equations. By using methods like substitution or elimination, we can determine these intersection points effectively.

Substitution Method

The substitution method is a powerful technique for solving systems of equations. It involves solving one of the equations for one variable in terms of the others and then substituting this expression into the remaining equations.
In our example, we begin by expressing \(y\) in terms of \(x\) using the simpler equation \(x + y = 1\). By doing so, we get: \(y = 1 - x\).
This step simplifies the problem because we can now substitute \(1 - x\) for \(y\) in the more complex equation \(A x + B y = C\). Through substitution, the equation transforms to \(A x + B(1 - x) = C\), and we now have a single equation with one variable, \(x\).
By solving for \(x\), we find its value directly. Afterwards, we substitute the newly found value back into the expression \(y = 1 - x\) to discover the value of \(y\). Consequently, the substitution method helps break down complex systems into more manageable calculations.