Problem 42 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: Analyze the System of Equations

We have two equations: \( ax + by = 0 \) and \( a^2x + b^2y = 1 \). Our goal is to find expressions for \(x\) and \(y\) in terms of \(a\) and \(b\).

Step 2: Solve for One Variable

From the first equation \( ax + by = 0 \), solve for \(x\): \( x = -\frac{b}{a}y \). Substitute this expression for \(x\) into the second equation.

Step 3: Substitute into the Second Equation

Substitute \(x = -\frac{b}{a}y\) into the second equation: \( a^2(-\frac{b}{a}y) + b^2y = 1 \) which simplifies to \( -ab y + b^2 y = 1 \).

Step 4: Combine Like Terms

Combine the terms: \( (b^2 - ab)y = 1 \). This can be factored further to: \( (b)(b-a)y = 1 \).

Step 5: Solve for y

Since \((b)(b-a)y = 1\), we solve for \(y\): \( y = \frac{1}{b(b-a)} \).

Step 6: Back-Substitute to Find x

Now that we have \(y\), substitute \(y = \frac{1}{b(b-a)}\) back into \(x = -\frac{b}{a}y\): \( x = -\frac{b}{a} \times \frac{1}{b(b-a)} = -\frac{1}{a(b-a)} \).

Step 7: Final Solution

The solutions are \( x = -\frac{1}{a(b-a)} \) and \( y = \frac{1}{b(b-a)} \).

Key Concepts

Variables Substitution

Variables substitution is a powerful approach to solving systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation. This helps in reducing the number of variables and simplifying the problem.
In this specific exercise, we started with the equations:

• \( ax + by = 0 \)
• \( a^2 x + b^2 y = 1 \)

The first step was to express \(x\) in terms of \(y\) using the first equation. By isolating \(x\), we derived:

• \( x = -\frac{b}{a} y \)

This expression for \(x\) was then substituted into the second equation. Substitution is often used when one equation is simpler to manipulate than the others. It allows us to systematically eliminate variables and work towards finding a solution. Remember, substitution is not just about plugging numbers; it’s about replacing entire expressions to better simplify and solve the problem.

Linear Equations

Linear equations form the backbone of solving systems like the one in this problem. They are equations that make a straight line when graphed, and typically have variables raised only to the power of one. In this exercise, we deal with a system of two linear equations:

• \( ax + by = 0 \)
• \( a^2 x + b^2 y = 1 \)

The system is linear because each variable appears only to the first power. The method of solving these equations often involves simplifying the system by dealing with one equation at a time. In our case, one equation was solved for a single variable, which made it possible to substitute into the other equation.
Linear equations are straightforward in structure, beneficial unless the equations turn out to be parallel, as no single solution exists in such scenarios. Here, by substituting and rearranging terms, we were able to maintain the linearity of the system and eventually find expressions for both \(x\) and \(y\).

Solution Methods

There are several solution methods for systems of equations, but variables substitution, as we used, is particularly effective when the equations are simple enough to manipulate easily. Other methods include graphing, elimination, and matrix operations.
The end goal is to find the values of the variables that satisfy both equations simultaneously. The key steps in our exercise included:

• Solving for one variable in terms of the others.
• Substituting back into the remaining equation.
• Reorganizing and simplifying to find a clear solution.

After substitution and simplification, we derived:

• \( y = \frac{1}{b(b-a)} \)
• \( x = -\frac{1}{a(b-a)} \)

These results give us the solutions to the system based on the substitution method uniquely fitting for this type of linear system. Each method serves its purpose, but substitution remains a simple yet powerful tool when applied to problems that lend themselves to this technique.