Problem 53 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understanding the Exercise

We are given a system of equations with two equations. Our goal is to solve for the variables \(x\) and \(y\) in terms of \(a\) and \(b\). The equations are: \(x + y = 0\) and \(x + ay = 1\).

Step 2: Substituting from the First Equation

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

Step 3: Substitute in the Second Equation

Substitute \(x = -y\) into the second equation \(x + ay = 1\):\[-y + ay = 1\].

Step 4: Simplifying the Second Equation

Combine like terms in the equation \(-y + ay = 1\): \(a = 1\), \ thus simplifying to \((a - 1)y = 1\).

Step 5: Solve for y

We solve for \(y\) by dividing both sides by \((a - 1)\), giving:\[y = \frac{1}{a - 1}\].

Step 6: Substitute y Back to Find x

Use the expression for \(y\) from the previous step and substitute it back into \(x = -y\):\[x = -\frac{1}{a - 1}\].

Step 7: Presenting the Solution

The solutions are \(x = -\frac{1}{a - 1}\) and \(y = \frac{1}{a - 1}\). These expressions give \(x\) and \(y\) in terms of \(a\).

Key Concepts

Substitution Method

The substitution method is a powerful and widely-used technique for solving systems of equations. This approach is especially useful when you have linear equations, as seen in the exercise. The core idea is to solve one of the equations for one variable and then substitute this expression into the other equation to find the second variable.

Let's break it down:

• Choose one equation and solve it for one of the variables. In the exercise, we used the first equation, x + y = 0, to express x as -y.
• Substitute the expression obtained into the other equation. Here, substituting x = -y into x + ay = 1 transformed the equation into -y + ay = 1.
• Now solve the new equation. This step typically simplifies the process by reducing the problem to a single variable equation.

The substitution method not only provides a systematic way to approach solving equations but also enhances your understanding of the relationship between variables. By manipulating the equations, you essentially peel back unnecessary layers, simplifying the problem at hand.

Solving Equations

Solving equations involves finding the values of variables that make the equation true. In mathematics, it's a process of determining the unknown variables that satisfy the conditions specified by the equations.

In our provided exercise, we have a system of two linear equations:

• First equation: x + y = 0
• Second equation: x + ay = 1

The target is to determine values for x and y in terms of a and b.

Each step in solving these equations aims at simplifying and manipulating the expressions to isolate the desired variables:

• Rearranging terms appropriately to get a variable by itself.
• Performing operations like addition, subtraction, multiplication, or division as needed.
• Ensuring each step logically follows the last, maintaining the equality.

This logical sequencing is crucial for problem-solving, allowing for complete solutions that satisfy both equations in the system.

Linear Equations

Linear equations are equations between two variables that produce straight lines when graphed. The general form of a linear equation in a two-variable context is expressed as ax + by = c, where a, b, and c are constants.

In the given exercise, we encountered two linear equations:

• Equation 1: x + y = 0
• Equation 2: x + ay = 1

These belong to the simplest category of equations, known as linear equations, because they graph as straight lines.

Key characteristics of linear equations include:

• Simplistic, non-exponential relationship between variables.
• Graphing results in straight lines.
• Easy to solve using algebraic methods, such as substitution or elimination.

Linear equations, like those in our problem, are foundational in algebra. They often serve as a stepping stone to more complex concepts, making understanding their properties and solutions crucial for advancing in mathematics.