Problem 54 - 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 \(A E \neq B D .\) $$ \begin{array}{r}{A x+B y=C} \\ {x+y=1}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the System of Equations

The problem presents a system of two linear equations in terms of the variables \(x\) and \(y\). The equations are: \(Ax + By = C\) and \(x + y = 1\). We are to solve for \(x\) and \(y\) using these equations.

Step 2: Solve the Second Equation for One Variable

Start with the simpler equation, \(x + y = 1\), and solve for one of the variables, say \(y\). We have:\[y = 1 - x\].

Step 3: Substitute into the First Equation

Substitute the expression for \(y\) from Step 2 into the first equation \(Ax + By = C\):\[A x + B(1 - x) = C\].

Step 4: Simplify and Solve for x

Expand and simplify the equation from Step 3:\[A x + B - B x = C\].Combine like terms:\[(A - B)x = C - B\].Solve for \(x\):\[x = \frac{C - B}{A - B}\].

Step 5: Substitute Back to Solve for y

Use the value of \(x\) from Step 4 in the equation \(y = 1 - x\):\[y = 1 - \frac{C - B}{A - B}\].

Step 6: Simplify the Expression for y

To simplify \(y\), rewrite it as:\[y = \frac{A - B - (C - B)}{A - B}\]Simplify further:\[y = \frac{A - C}{A - B}\].

Key Concepts

Linear Algebra

Linear algebra is a cornerstone of modern mathematics. It deals extensively with vector spaces and linear mappings, providing tools to handle systems of linear equations effectively. This area is fundamental in numerous fields, including engineering, physics, computer science, and more. In the context of solving systems of equations, linear algebra provides methodologies to simplify and resolve problems. In simpler terms, linear algebra examines lines, planes, and how they interact in multiple dimensions. This is particularly important when you want to find common solutions to multiple equations - known as the solution to a system of equations. Whether dealing with two or more equations, the aim is always to find the point(s) satisfying all equations within the system.

Substitution Method

The substitution method is a popular technique used to solve systems of linear equations. It is particularly effective when dealing with equations that are easily rearranged to express one variable in terms of another. This method reduces the complexity by transforming a system into a single-variable equation. Here's a simple outline:

• Solve one of the equations for one variable. This helps to express one variable explicitly in terms of the other(s).
• Substitute the expression obtained into the other equation(s). This substitution eliminates one variable, making it easier to solve for the remaining unknowns.
• Solve the new equation formed. The result gives you the value for one variable, which can then be back-substituted to find other variable(s).

By following these steps, you convert a complicated system of equations into more manageable parts, leading to an exact solution.

Solving Equations

Solving equations is fundamental to mathematics, especially when working with systems of linear equations. The objective is to find values for the variables that satisfy all the given equations simultaneously. This entails several logical steps including simplifying equations and rearranging terms. Key steps usually include:

• Simplifying expressions by combining like terms and using distributive properties.
• Isolating variables to one side of the equation. This step can often involve arithmetic operations such as addition, subtraction, multiplication, or division.
• Substituting known values or expressions to limit the number of unknowns, which clarifies the solution path.

The aim is to reach a point where each equation is balanced and can display the unknown variable(s). In complex systems, solving requires patience, precision, and sometimes iterative checking to ensure correctness.