Problem 56 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

\( x = \frac{1}{a^2 - ab} \), \( y = -\frac{a}{b(a^2 - ab)} \).

Step 1: Use Substitution for Elimination

First, from the equation \( ax + by = 0 \), express \( y \) in terms of \( x \). Solving for \( y \) gives \( y = -\frac{a}{b}x \).

Step 2: Substitute into the Second Equation

Substitute \( y = -\frac{a}{b}x \) into \( a^2x + b^2y = 1 \). This gives us \( a^2x + b^2(-\frac{a}{b}x) = 1 \). Simplify to get \( a^2x - abx = 1 \).

Step 3: Solve for x

Factor \( x \) out of the equation \( a^2x - abx = 1 \), which simplifies to \( x(a^2 - ab) = 1 \). Therefore, \( x = \frac{1}{a^2 - ab} \).

Step 4: Solve for y Using x

Substitute \( x = \frac{1}{a^2 - ab} \) back into the expression for \( y \) that we found in Step 1: \( y = -\frac{a}{b} \times \frac{1}{a^2 - ab} \). Simplifying gives \( y = -\frac{a}{b(a^2 - ab)} \).

Key Concepts

Substitution Method

The substitution method is a fundamental technique used to solve systems of simultaneous equations. The core idea is to solve one of the equations for one variable and then substitute that expression back into the other equation. This way, you reduce the system to a single equation with one unknown, making it easier to solve.

In this exercise, we start with the equation \( ax + by = 0 \). By rearranging this equation, we express \( y \) in terms of \( x \), yielding \( y = -\frac{a}{b}x \).

Now, we take this expression for \( y \) and substitute it into the second equation \( a^2x + b^2y = 1 \). This substitution changes the equation into \( a^2x + b^2(-\frac{a}{b}x) = 1 \). Simplifying the expression gives us a direct equation in terms of \( x \).

One major benefit of the substitution method is that it allows us to focus on just one variable at a time by transforming the original system of equations. This makes the problem more manageable, especially when it involves algebraic expressions.

Elimination Method

The elimination method is another popular strategy to tackle systems of simultaneous equations. It involves manipulating the equations such that adding or subtracting them eliminates one of the variables.

In this task, while we primarily use substitution, elimination is implicit in how we handle the equations. Each equation is manipulated to isolate terms, which effectively "eliminates" one variable in the sense that we eventually solve for a single variable at a time.

Elimination is particularly efficient when the coefficients of one of the variables are already balanced between the equations, allowing for straightforward addition or subtraction to cancel that variable. Although this exercise does not explicitly focus on traditional elimination, understanding both methods provides flexibility in problem-solving.

Algebraic Expressions

Algebraic expressions are at the heart of solving simultaneous equations. They consist of numbers, variables, and arithmetic operations. In problems like this, comprehension of algebraic manipulation is crucial.

For example, expressions such as \( ax + by = 0 \) involve variables \( x \) and \( y \), which need to be resolved with respect to constants \( a \) and \( b \).

Through the entire problem-solving process, simplifying algebraic expressions is fundamental. Performing operations within these expressions requires precision:

• Combining like terms, as seen in \( a^2x - abx \).
• Factoring out common terms, which is used when solving \( x(a^2 - ab) = 1 \).
• Simplifying fractions, such as \( y = -\frac{a}{b(a^2 - ab)} \).

Mastering these manipulations ensures that you can effectively tackle simultaneous equations and other algebra problems.