Problem 39 - Algebra and Trigonometry

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

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

Step 1: Analyze the Problem

We are given a system of equations with two variables and two equations. The equations are: \(x + y = 0\) and \(x + ay = 1\). We need to express \(x\) and \(y\) in terms of the parameters \(a\) and \(b\).

Step 2: Solve Equation for y

From the first equation, \(x + y = 0\), we can express \(y\) in terms of \(x\): \(y = -x\).

Step 3: Substitute y in Second Equation

Substitute \(y = -x\) from Step 2 into the second equation \(x + ay = 1\). This gives us: \(x + a(-x) = 1\).

Step 4: Simplify and Solve for x

Simplify the equation \(x + a(-x) = 1\) to \(x - ax = 1\). Factor out \(x\) to get \((1 - a)x = 1\). Solve for \(x\) by dividing both sides by \(1 - a\), resulting in \(x = \frac{1}{1-a}\).

Step 5: Substitute x back to find y

Substitute \(x = \frac{1}{1-a}\) into \(y = -x\) from Step 2. This results in \(y = -\frac{1}{1-a}\).

Step 6: State the Solution

We have found expressions for \(x\) and \(y\) in terms of the parameter \(a\): \(x = \frac{1}{1-a}\) and \(y = -\frac{1}{1-a}\).

Key Concepts

Solving Linear Equations

In algebra, solving linear equations involves finding the values of unknown variables that satisfy all equations in a given system. Linear equations are called so because they graph as straight lines when plotted on a coordinate plane.

When dealing with a system of equations, a solution is a set of values for the variables that makes both equations true at the same time. For example, in our problem, we have to discover the values of \( x \) and \( y \) that satisfy both equations \( x + y = 0 \) and \( x + ay = 1 \).

Solving this can be achieved by methods such as substitution or elimination, where the goal is to isolate each variable. A good approach starts with pondering which method will provide a quicker route to isolating a variable with less complexity.

Substitution Method

The substitution method is a technique used to solve a system of equations where one equation is solved for one variable and this expression is used in another equation. This method allows us to work with simpler equations that make the process of finding the solution less cumbersome.

In our exercise, we applied the substitution method by first isolating \( y \) in the equation \( x + y = 0 \), yielding \( y = -x \).
Substituting \( -x \) for \( y \) in the second equation \( x + ay = 1 \), transforms it into an equation with only one variable.

• This substitution modifies the equation into \( x + a(-x) = 1 \), which simplifies to \( x - ax = 1 \).
• Through this process, we smartly reduce the system into equations manageable with basic algebraic techniques.

Substitution is often preferred when it is easy to express one variable in terms of another, as was the case here.

Algebraic Expressions

An algebraic expression consists of numbers, variables, and arithmetic operations. These expressions can represent a variety of mathematical relationships and allow us to form equations that describe real-world situations.

In our system of equations, \( x + y = 0 \) and \( x + ay = 1 \), the expressions are composed of variables \( x \) and \( y \), constants, and the arithmetic operations of addition and multiplication. Emphasizing the role and structure of such expressions is crucial in understanding how they are manipulated to find variables’ values.

• Understanding the core structure of an algebraic equation involves recognizing the like terms and coefficients and knowing how to efficiently rearrange and simplify expressions.
• In this context, the expression \( x - ax = 1 \) simplifies to \( (1-a)x = 1 \), illustrating a key manipulation - factoring out a common factor \( x \).

These skills are vital for solving not only simple equations but more complex ones that are encountered as we dig deeper into algebra.