Problem 55 - Precalculus Mathematics for Calculus

Question

Find \(x\) and \(y\) in terms of \(a\) and \(b\). $$\left\\{\begin{array}{l}a x+b y=1 \\\b x+a y=1\end{array}\left(a^{2}-b^{2} \neq 0\right)\right.$$

Step-by-Step Solution

Answer

\( x = \frac{a - b}{a^2 - b^2} \) and \( y = \frac{a - b}{a^2 - b^2} \).

Step 1: Understand the System of Equations

We are given two equations:1. \( ax + by = 1 \)2. \( bx + ay = 1 \)Our goal is to solve for \( x \) and \( y \) in terms of \( a \) and \( b \).

Step 2: Express One Variable in Terms of the Other

Let's express \( x \) from the first equation, i.e., \( ax + by = 1 \):\[ x = \frac{1 - by}{a} \]Similarly, express \( x \) from the second equation, \( bx + ay = 1 \):\[ x = \frac{1 - ay}{b} \]

Step 3: Equate Expressions for x

Equate the two expressions for \( x \):\[ \frac{1 - by}{a} = \frac{1 - ay}{b} \]By cross-multiplying, we will have:\[ b(1 - by) = a(1 - ay) \]

Step 4: Simplify and Solve for y

Expand and simplify the equation:\[ b - b^2y = a - a^2y \]Rearranging for \( y \):\[ a^2y - b^2y = a - b \]\[ (a^2 - b^2)y = a - b \]

Step 5: Determine y

Solving for \( y \):\[ y = \frac{a - b}{a^2 - b^2} \]

Step 6: Substitute to Find x

Substitute \( y = \frac{a - b}{a^2 - b^2} \) back into the expression for \( x \):\[ x = \frac{1 - b\left(\frac{a - b}{a^2 - b^2}\right)}{a} \]Simplifying:\[ x = \frac{a^2 - b^2 - b(a - b)}{a(a^2 - b^2)} \]\[ x = \frac{a^2 - b^2 - ab + b^2}{a(a^2 - b^2)} \]\[ x = \frac{a^2 - ab}{a(a^2 - b^2)} \]\[ x = \frac{a - b}{a^2 - b^2} \]

Step 7: Final Values

We find that \( x = \frac{a - b}{a^2 - b^2} \) and \( y = \frac{a - b}{a^2 - b^2} \), ensuring they are consistent with respect to both equations.

Key Concepts

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate and solve for a particular variable. In solving systems of linear equations, this process is essential for expressing one variable in terms of another. Within the given problem, this technique is used to express \(x\) from both equations separately:

• From the first equation, \(x\) is expressed as \( x = \frac{1 - by}{a} \).
• From the second equation, \(x\) is expressed as \( x = \frac{1 - ay}{b} \).

This manipulation is key because it sets up the system for further operations, allowing us to use substitution or elimination strategies. By changing the forms of our equations, we create avenues for simplification, which gets us one step closer to finding the solutions \(x\) and \(y\) in terms of \(a\) and \(b\). Remember, delicate handling of terms and careful algebraic steps are crucial to avoid errors during manipulation.

Solving Linear Equations

Solving linear equations involves finding the values of the variables that satisfy all equations in the system. Once we've manipulated the equations, the next step is to solve by equating terms. In our problem:

• We equate the two expressions for \(x\): \( \frac{1 - by}{a} = \frac{1 - ay}{b} \).
• This equation is then simplified through cross-multiplication.

This stage reduces the problem to a single equation in one variable, \(y\). Subsequently, rearrangement leads to the form \((a^2 - b^2)y = a - b\). Solving this linear equation for \(y\), we found \(y = \frac{a - b}{a^2 - b^2}\). Once \(y\) is determined, its value can be substituted back into any original equation to solve for \(x\). Therefore, each step methodically breaks the problem down, ensuring that we respect the rules of equality and logic to find the variables accurately.

Cross-Multiplication

Cross-multiplication is a technique commonly used when we have a proportion or equation involving fractions. This method simplifies the equation by eliminating the denominators. In the context of our exercise, after expressing \(x\) in terms of \(y\) from both equations, cross-multiplication is applied:

• We had \( \frac{1 - by}{a} = \frac{1 - ay}{b} \).
• Cross-multiplying, \(b(1 - by) = a(1 - ay)\), removes the fractions and leads to a simpler equation that can be easily solved.

During cross-multiplication, always ensure to distribute terms correctly to avoid mistakes in expansion. This technique turns what might appear a complex fractional equation into a clearer, linear format, paving the way for straightforward algebraic manipulation. By applying cross-multiplication accurately, we further simplify the conditions under which \(x\) and \(y\) are solved, providing a clear path to the final solution.