Problem 58 - Precalculus: Mathematics for Calculus

Question

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

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

Step 1: Use Elimination to Solve for One Variable

To eliminate one of the variables, we can manipulate the given equations. Start with the first equation, \( ax + by = 0 \). This can be rewritten as \( ax = -by \), so \( x = -\frac{b}{a}y \). Substituting \( x = -\frac{b}{a}y \) into the second equation \( a^2x + b^2y = 1 \) will help find \( y \) in terms of \( a \) and \( b \).

Step 2: Substitute and Simplify

Substitute \( x = -\frac{b}{a}y \) into \( a^2x + b^2y = 1 \) to get:\[a^2\left(-\frac{b}{a}y\right) + b^2y = 1\]Simplify the equation:\[-abyy + b^2y = 1\]Factor out \( y \):\[(-ab + b^2)y = 1\]

Step 3: Solve for y

Solving the equation \((-ab + b^2)y = 1 \) for \( y \) gives:\[y = \frac{1}{b^2 - ab}\]This represents \( y \) in terms of \( a \) and \( b \).

Step 4: Substitute Back to Find x

Substitute \( y = \frac{1}{b^2 - ab} \) back into \( x = -\frac{b}{a}y \). This results in:\[x = -\frac{b}{a}\left(\frac{1}{b^2 - ab}\right)\]Simplifying gives:\[x = \frac{-b}{a(b^2 - ab)}\]

Step 5: Final Answer in Terms of a and b

The solutions for \( x \) and \( y \) in terms of \( a \) and \( b \) are:\[x = \frac{-b}{a(b^2 - ab)}\]\[y = \frac{1}{b^2 - ab}\]

Key Concepts

Elimination Method

The elimination method is a technique used to solve systems of linear equations. It involves manipulating the equations in such a way that one of the variables cancels out, allowing us to solve for the remaining variable. In the given exercise, to eliminate the variable \( x \), we started with the equation \( ax + by = 0 \). By rewriting this as \( ax = -by \), we can express \( x \) in terms of \( y \), specifically \( x = -\frac{b}{a}y \).
This step is crucial because it allows us to substitute into the second equation, \( a^2x + b^2y = 1 \), effectively reducing the number of variables in the equation. The elimination method not only simplifies the process but also clarifies the relationships between the variables. It’s like reducing clutter, making it easier to see what you’re truly dealing with.

Substitution Method

The substitution method involves solving one of the equations for one variable and then substituting this expression into the other equation. After using the elimination method to express \( x \) as \( x = -\frac{b}{a}y \), this expression was substituted into the second equation \( a^2 x + b^2 y = 1 \).
After substitution, the equation becomes \( a^2 \left( -\frac{b}{a}y \right) + b^2y = 1 \), which simplifies to the form \( (-ab + b^2)y = 1 \).
This step simplifies solving the system into finding one variable at a time. Substitution is a valuable method when one equation is already solved for one variable, or can be easily manipulated to be so. It gives a clear pathway to solve the problem incrementally.

Solving Linear Equations

Solving linear equations means finding the values of the variables that make the equations true. The ultimate goal of both the elimination and substitution methods is to arrive at solutions for \( x \) and \( y \).

• Starting with the equation \( (-ab + b^2)y = 1 \), solving for \( y \) gives \( y = \frac{1}{b^2 - ab} \).
• Substituting \( y \) back to find \( x \), as in \( x = -\frac{b}{a}\left(\frac{1}{b^2 - ab}\right) \), simplifies to \( x = \frac{-b}{a(b^2 - ab)} \).

Each step in solving linear equations builds on the previous step, leading us to these solutions. It’s like piecing together a puzzle—each method plays a critical role in uncovering the pieces until the complete picture is visible. In this exercise, using these techniques, we transformed the original equations into understandable solutions for any \( a \) and \( b \) as long as \( a eq 0 \) and \( b eq 0 \).