Problem 42 - Precalculus: Mathematical for Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Express y in terms of x from the first equation

Start with the equation \( ax + by = 0 \). Solve for \( y \) by isolating it on one side: \( by = -ax \). Therefore, \( y = -\frac{a}{b}x \).

Step 2: Substitute y in the second equation

Substitute \( y = -\frac{a}{b}x \) into the second equation \( a^2x + b^2y = 1 \) to get \( a^2x + b^2(-\frac{a}{b}x) = 1 \). Simplify to get \( a^2x - abx = 1 \).

Step 3: Simplify to solve for x

Factor out \( x \) from \( a^2x - abx = 1 \) to get \( x(a^2 - ab) = 1 \). Solve for \( x \), which yields \( x = \frac{1}{a^2-ab} \).

Step 4: Solve for y using y = -a/b x

Use \( x = \frac{1}{a^2-ab} \) to find \( y \). Substitute \( x \) into \( y = -\frac{a}{b}x \), resulting in \( y = -\frac{a}{b} \cdot \frac{1}{a^2-ab} = -\frac{a}{b(a^2-ab)} \). Simplify to get \( y = -\frac{a}{a^2b - b^2a} = \frac{-a}{b(a^2-ab)} \).

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a crucial skill in solving systems of linear equations. It involves rearranging equations to simplify and isolate variables. In our exercise, the first step involves manipulating the equation \( ax + by = 0 \) to express \( y \) in terms of \( x \). By isolating \( y \), it can be seen that \( y = -\frac{a}{b}x \). This is done by performing the same operation on both sides of the equation, ensuring the balance and equality are maintained.

Key steps in algebraic manipulation:

• Identify which variable to isolate based on the requirements of the problem.
• Perform operations equally on both sides of the equation to maintain equality.
• Simplify the expression to make substitution easier in subsequent steps.

Understanding algebraic manipulation helps in transforming equations into more workable formats, which is essential when dealing with complicated systems.

Substitution Method

The substitution method is a strategy to solve systems of linear equations by replacing one variable in one equation with an expression derived from another equation. In this exercise, after expressing \( y \) as \( -\frac{a}{b}x \), we substitute this expression into the second equation \( a^2x + b^2y = 1 \). This results in a single equation in terms of \( x \) only: \[ a^2x + b^2\left(-\frac{a}{b}x\right) = 1 \]

This substitution simplifies the problem to an equation with only one variable, making it easier to solve. Here's why substitution is useful:

• Simplifies complex systems by reducing the number of variables.
• Helps in methodically approaching the solution.
• Efficient when one equation is easily solvable for a variable.

The substitution method is powerful for tackling linear equations systematically, paving the way for finding precise solutions.

Solving Linear Equations

Solving linear equations is the final step to find the values of variables. After we substitute \( y \) and simplify, we get the equation: \[ a^2x - abx = 1 \] Factoring out \( x \), we have: \[ x(a^2 - ab) = 1 \]
By solving for \( x \), we find: \( x = \frac{1}{a^2-ab} \).

Next, we substitute this value of \( x \) back into the expression for \( y \), \( y = -\frac{a}{b}x \), to obtain: \( y = \frac{-a}{b(a^2-ab)} \).

Some key points about solving linear equations include:

• Always perform inverse operations to isolate the variable.
• Check your work by plugging the solution back into the original equations to verify.
• Factorization can be an effective tool for simplifying the equations.

By the end of this process, each variable will be expressed clearly in terms of the given parameters, providing a comprehensive solution to the system.