Problem 55 - Precalculus: Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Express y in terms of x

From the equation \( x + y = 0 \), we can isolate \( y \) by subtracting \( x \) from both sides: \( y = -x \).

Step 2: Substitute y in the second equation

Take the expression for \( y = -x \) and substitute it into the second equation \( x + a y = 1 \). This gives us: \( x + a(-x) = 1 \) or more simply \( x - ax = 1 \).

Step 3: Solve for x

Factor \( x \) out from the equation \( x - ax = 1 \), which gives: \( x(1 - a) = 1 \). Since \( a eq 1 \), divide both sides by \( 1-a \): \( x = \frac{1}{1-a} \).

Step 4: Substitute x back to find y

Now substitute \( x = \frac{1}{1-a} \) into \( y = -x \) to find \( y \): \( y = -\frac{1}{1-a} \).

Key Concepts

Linear Equations

Linear equations form the backbone of algebra, dealing with expressions where variables are raised only to the first power. In the simultaneous equations given, each equation is linear. The first equation, \( x + y = 0 \), suggests that the sum of \( x \) and \( y \) equals zero. This is a classic linear equation where you can quickly find one variable if the other is known.

The second equation, \( x + ay = 1 \), represents another linear relationship between \( x \) and \( y \). These types of equations can be graphically represented as straight lines on a coordinate plane. Where these lines intersect gives the solution to the system, providing the values for both \( x \) and \( y \).
Understanding how to manipulate and solve these equations is fundamental. They allow us to model real-world problems and find values of interest, resulting in the intersection of our defined linear relationships.

Algebraic Manipulation

Algebraic manipulation is a key skill needed to solve linear equations, especially when dealing with simultaneous equations. This involves taking the original equations and transforming them in order to isolate one variable, making the equation easier to solve.

In the given problem, we used algebraic manipulation right from the start. From the equation \( x + y = 0 \), we isolated \( y \) by subtracting \( x \) from both sides, giving us \( y = -x \). This simple rearrangement allows us to substitute \( y \) in the second equation without any complex operations.

Further manipulation was used in the second equation, where substituting \( y = -x \) into \( x + ay = 1 \) gave us \( x - ax = 1 \). Factorization allowed us to isolate \( x \), and dividing both sides by \( 1-a \) gave the result for \( x \). Breaking down complex equations into simpler forms is the underlying essence of algebraic manipulation. It's the way to transform and solve equations effectively, especially when dealing with variables and unknowns.

System of Equations

A system of equations is a set of two or more equations that work together to find a common solution. In the exercise, the goal was to find values for \( x \) and \( y \) that satisfy both equations: \( x + y = 0 \) and \( x + a y = 1 \).

To solve the system, we need to ensure each equation is considered, and the variables are found to fulfill both expressions simultaneously. Systems of equations can often be solved using different methods:

• Substitution, where you solve one equation for one variable and substitute that into the other.
• Elimination, which involves adding or subtracting equations to cancel out one variable, making it easier to solve for the other.

In this problem, substitution was used. After expressing \( y \) in terms of \( x \) from the first equation, this expression was plugged into the second equation, allowing us to solve for \( x \). Once \( x \) was found, it was substituted back to find \( y \). Systems of equations are crucial in math and science, providing a framework to solve problems where multiple conditions or constraints are present.