Problem 55 - Algebra and Trigonometry

Question

\(53-56\) Find \(x\) and \(y\) in terms of \(a\) and \(b\). $$ \left\\{\begin{array}{l}{a x+b y=1} \\ {b x+a y=1}\end{array} \quad\left(a^{2}-b^{2} \neq 0\right)\right. $$

Step-by-Step Solution

Answer

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

Step 1: First Equation Substitution

We have the first equation \( ax + by = 1 \). We can express one variable in terms of the other. Let's express \( x \) in terms of \( y \): \( x = \frac{1 - by}{a} \).

Step 2: Substitute in the Second Equation

Substitute \( x = \frac{1 - by}{a} \) into the second equation \( bx + ay = 1 \). This gives us: \[ b\left(\frac{1 - by}{a}\right) + ay = 1 \]. Simplifying, we get: \[ \frac{b - b^2y}{a} + ay = 1 \].

Step 3: Clear the Denominator

Multiply each term by \( a \) to clear the fraction: \( b - b^2y + a^2y = a \). Rearranging terms gives: \( (a^2 - b^2)y = a - b \).

Step 4: Solve for y

Since \( a^2 - b^2 eq 0 \), we can solve for \( y \): \( y = \frac{a - b}{a^2 - b^2} \).

Step 5: Solve for x

Substitute \( y = \frac{a - b}{a^2 - b^2} \) back into \( x = \frac{1 - by}{a} \). We get: \( x = \frac{1 - b\left(\frac{a - b}{a^2 - b^2}\right)}{a} \). Simplifying the expression for \( x \), we find: \( x = \frac{b - a}{b^2 - a^2} = \frac{b - a}{a^2 - b^2} \).

Step 6: Final Expressions for x and y

We have found the expressions: \( x = \frac{b - a}{a^2 - b^2} \) and \( y = \frac{a - b}{a^2 - b^2} \). These are the values of \( x \) and \( y \) in terms of \( a \) and \( b \).

Key Concepts

Substitution Method

The substitution method is a powerful technique used to solve simultaneous equations. It involves expressing one variable in terms of the other and substituting this expression into another equation. This method can simplify complex algebraic problems, making them more manageable.

• Start by solving one of the equations for one of the variables. For instance, if working with the equation \( ax + by = 1 \), solve for \( x \): \( x = \frac{1 - by}{a} \).
• Next, take this expression and substitute it into the other equation. This will result in an equation with a single variable, simplifying the problem.

By using substitution, you reduce the problem to a sequence of simpler tasks. This method is particularly useful when the equations are interdependent, like in our original exercise.

System of Linear Equations

A system of linear equations consists of two or more linear equations involving the same set of variables. Solving these systems means finding the values of the variables that satisfy all equations simultaneously.
Simultaneous equations can represent real-world problems where multiple conditions must be satisfied at once. For example:

• The equations \( ax + by = 1 \) and \( bx + ay = 1 \) in the exercise are a system of linear equations.
• Each equation represents a line in coordinate geometry. The solution corresponds to the intersection point of these lines.

Understanding systems of equations helps in various fields such as engineering, physics, and economics, where quantitative relationships between variables are analyzed.

Solving Algebraic Equations

Solving algebraic equations involves finding the values of variables that make the equations true. This process can entail different methods, including substitution, elimination, and graphing.
In our exercise, solving for \( y \) and \( x \) included:

• First, isolating \( y \) by rearranging terms after substitution, we get the equation \( (a^2 - b^2)y = a - b \).
• Solving for \( y \) gives \( y = \frac{a - b}{a^2 - b^2} \), assuming \( a^2 - b^2 eq 0 \) ensures the divisor isn't zero.
• Substituting the value of \( y \) back to find \( x \), we ended up with the expression \( x = \frac{b - a}{a^2 - b^2} \).

Being proficient in solving algebraic equations is essential across various math-based disciplines, as it forms the foundation for more advanced problem-solving.