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

Step-by-Step Solution

Answer

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

Step 1: Analyze the System of Equations

The given system of equations is:\[ \begin{align*} A x + y &= 0 \ B x + y &= 1 \end{align*} \] We have a pair of linear equations in two variables, \(x\) and \(y\). Our task is to solve for \(x\) and \(y\) in terms of the parameters \(A\) and \(B\).

Step 2: Eliminate y by Subtraction

To eliminate \(y\), subtract the first equation from the second. This gives:\[ Bx + y - (Ax + y) = 1 - 0 \] Simplifying, we find:\[ (B - A)x = 1 \]

Step 3: Solve for x

With the equation \((B - A)x = 1\), solve for \(x\) by dividing both sides by \((B - A)\):\[ x = \frac{1}{B - A} \] Provided \(A eq B\), this operation is valid.

Step 4: Solve for y using the First Equation

Substitute the value of \(x\) back into the first equation: \(Ax + y = 0\). This becomes:\[ A\left(\frac{1}{B - A}\right) + y = 0 \] Simplifying this expression gives:\[ \frac{A}{B - A} + y = 0 \]

Step 5: Isolate y

Isolate \(y\) by subtracting \(\frac{A}{B - A}\) from both sides of the equation:\[ y = -\frac{A}{B - A} \]

Step 6: Express Solution in Terms of Given Variables

The solutions for \(x\) and \(y\) in terms of \(A\) and \(B\) are:\[ x = \frac{1}{B - A} \quad \text{and} \quad y = -\frac{A}{B - A} \] This completes the solution in terms of the given variables.

Key Concepts

Linear Equations

Linear equations are mathematical expressions where each term is either a constant or the product of a constant and a single variable. These equations are represented in the form of straight lines when plotted on a graph. In our problem, we worked with two linear equations:

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

Such equations are "linear" because they generate straight lines if graphed. This term 'linear' derives from the Latin word "linearis," which means pertaining to or resembling a line.
In solving the provided system of equations, like the ones above, the goal is typically to find the values of unknown variables which satisfy all equations in the system. This involves understanding the relationships and intersections that occur between these linear lines.

Variables

Variables are symbols used in equations to stand for unknown values. In our system of equations (\(A x + y = 0 \)and\(B x + y = 1 \)), the variables are \(x\) and \(y\).
Variables can change and take on different values within equations, generally depending on the surrounding components like constants or coefficients. They are essential tools that represent an unknown number, allowing mathematicians to form expressions and equations that model real-world situations.Specifically, in our solution, we found specific expressions for these variables:

• Variable \(x\) was solved to be \(x = \frac{1}{B - A}\)
• Variable \(y\), similarly, turned out to be \(y = -\frac{A}{B - A}\)

These values of \(x\) and \(y\) demonstrate how variables take on specific values based on the parameters within the system.

Parameters

Parameters are constants in the equations that help describe the system but remain fixed during the process of solving the equations. In our exercise, the parameters are given by the letters \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\), with a primary focus on \(A\) and \(B\) for this particular solution.Unlike variables, parameters do not transform or change values within the confines of this specific equation-solving context. They essentially give context or boundary to the problem.
For the given problem:

• The operations and final expressions for \(x\) and \(y\), such as \(x = \frac{1}{B - A}\), were derived based significantly on these parameters \(A\) and \(B\).

Thus, parameters serve as the constants or fixed numbers that influence the behavior and solution of the equations at hand.

Solutions in terms of Variables

Solving a system of equations in terms of variables, involves expressing the unknown variables within the equations using the given parameters. In our provided solution, we derived the solutions for \(x\) and \(y\) as expressions involving the parameters \(A\) and \(B\).The steps entailed solving each variable independently and ensuring each is expressed as a function of these parameters, resulting in:

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

By expressing these solutions purely in terms of parameters, we can determine the values of \(x\) and \(y\) for any specified \(A\) and \(B\). This method highlights the power of using algebraic manipulation to make sense of complex systems and to find solutions in a generalized manner, applicable to many specific cases.