Problem 57 - 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: Identify the System of Equations

We have two simultaneous equations: 1. \( ax + by = 1 \)2. \( bx + ay = 1 \)We need to solve this system for \(x\) and \(y\) in terms of \(a\) and \(b\).

Step 2: Solve for x in terms of y

From the first equation, \( ax + by = 1 \), solve for \( x \):\[ x = \frac{1 - by}{a} \] if \(a eq 0\).

Step 3: Substitute x back into the second equation

Substitute \( x = \frac{1 - by}{a} \) into the second equation \( bx + ay = 1 \):\[ b\left(\frac{1 - by}{a}\right) + ay = 1 \]

Step 4: Simplify and Solve for y

Multiply through by \(a\) to eliminate fractions:\[ b(1 - by) + a^2y = a \]\[ b - b^2y + a^2y = a \] \[ y(a^2 - b^2) = a - b \]Now solve for \( y \): \[ y = \frac{a-b}{a^2-b^2} \]

Step 5: Back Substitute to find x

Substitute \( y = \frac{a-b}{a^2-b^2} \) into \( x = \frac{1 - by}{a} \):\[ x = \frac{1 - b\left(\frac{a-b}{a^2-b^2}\right)}{a} \]\[ x = \frac{a^2-b^2 - b(a-b)}{a(a^2-b^2)} \]Simplify:\[ x = \frac{a^2-b^2 - (ab - b^2)}{a(a^2-b^2)} \]\[ x = \frac{a^2 - ab}{a(a^2-b^2)} \]\[ x = \frac{a(a-b)}{a(a^2-b^2)} \]\[ x = \frac{a-b}{a^2-b^2} \]

Step 6: Verify the Solution

Verify by substituting \( x = \frac{a-b}{a^2-b^2} \) and \( y = \frac{a-b}{a^2-b^2} \) back in both original equations.This satisfies both equations since it results in equivalences for the right-hand sides.

Key Concepts

System of Linear Equations

Linear equations are equations of the first order. They create straight lines when graphed on a coordinate plane. A system of linear equations consists of two or more such equations. The primary goal when dealing with these systems is to find the values of the variables that satisfy all equations simultaneously.

In the given exercise, we have a system consisting of two linear equations:

• First equation: \( ax + by = 1 \)
• Second equation: \( bx + ay = 1 \)

To "solve" a system means to find the values of variables (in this case, \(x\) and \(y\)) that satisfy both equations together. The challenge here lies in finding a solution that works for the arranged set of variables and coefficients.

Solving Algebraically

When solving a system of linear equations algebraically, we don't rely on graphical or numerical methods, but rather manipulate the equations using algebraic steps to find the exact solution. This often involves methods like substitution or elimination.

In our solved example, we applied the substitution method. While it seems daunting at first, solving algebraically meticulously brings us closer to finding the solution without using guesswork. Each step builds on the last to, piece by piece, untangle the variables from their roles within the complex system of equations. Algebraic solutions are critical in scenarios where precise values are necessary or where graphical solutions might be too imprecise due to scale or resolution.

Substitution Method

The substitution method is a common strategy for solving systems of equations. We solve one of the equations for one variable in terms of the others and substitute this into the remaining equations.

Here's how it worked in the exercise:

• First, we solved the first equation for \(x\) in terms of \(y\): \( x = \frac{1-by}{a} \).
• Next, we substituted this expression for \(x\) into the second equation.

Through these substitutions, we reduce the simultaneous equations into a single equation with one variable, which simplifies the process significantly. The substitution method is particularly handy when one of the variables can be easily expressed in terms of the other.

Simplification in Algebra

Simplification is the heart of efficient algebra. It involves reducing expressions or equations to their simplest form, making them easier to work with and solve.

In the context of our solution, simplification arose at several key points:

• Eliminating fractions by multiplying through by a common denominator (specifically multiplying by \(a\) to clear the fraction).
• Combining like terms and factoring expressions to make the equation easier to solve.

Ultimately, simplification helps clarify complex expressions, uncovering the structure underlying mathematical problems and making it apparent what steps to take next. It transforms challenging equations into a more manageable format, paving the way for finding solutions.