Problem 23 - College Algebra Essentials

Question

Determine whether each equation defines \(y\) as a function of \(x .\) $$ x y+2 y=1 $$

Step-by-Step Solution

Answer

Yes, the given equation defines y as a function of x, as long as \(x ≠ -2\).

Step 1: Isolate y

The first step is to isolate y in the given equation. To do this, first factor y out of the left-side of the equation to get: \(y(x + 2) = 1\). Then divide everything by \(x + 2\) to isolate y: \(y = \frac{1}{x + 2}\).

Step 2: Check if the equation defines y as a function of x

The equation \(y = \frac{1}{x + 2}\) is a function because for each x-value, there is only one y-value. However, remember that \(x ≠ -2\) since the denominator of a fraction cannot be zero.

Key Concepts

Isolating variables

In mathematics, isolating variables is a fundamental skill when working with equations. The goal is to solve for one specific variable, making it the subject of the equation. This process is crucial when you want to express one variable explicitly in terms of others.
For instance, consider the equation given in our exercise: \( x y + 2 y = 1 \). Here we need to solve for \( y \). To do this, we first notice that \( y \) appears in both terms on the left side. Our first move is to factor out \( y \) from those terms.

• Factor out y: \( y(x + 2) = 1 \).
• Divide both sides by \( (x + 2) \) to solve for \( y \): \( y = \frac{1}{x + 2} \).

This successful isolation of \( y \) is an essential step towards understanding the relationship between \( x \) and \( y \) in the context of the equation.

Solving equations

Solving equations involves finding values for the variables that satisfy the equation. This process often requires multiple steps, depending on the complexity of the equation.
In our example, once we isolated \( y \), the resulting equation was \( y = \frac{1}{x + 2} \). This expression is in its simplest form and represents a solution where \( y \) is defined explicitly in terms of \( x \).
When we solve equations like these, we ask if the equation correctly defines one variable in terms of the other. Here, for every \( x \) (except \( x = -2 \)), there is a single corresponding \( y \). This establishes \( y \) as a function of \( x \).
Being a function means that for each input \( x \), we obtain exactly one output \( y \), which is checked and ensured in our solution.

Domain restrictions

In algebra, domain restrictions are essential when dealing with functions. They tell us which values are permissible for our variables. Domain issues often arise when fractions are involved, particularly where there's division by a variable expression.
In our exercise, the equation became \( y = \frac{1}{x + 2} \) after isolating \( y \). Here, the denominator \( x + 2 \) cannot be zero, because division by zero is undefined.

• Identify the restriction: \( x + 2 eq 0 \).
• Therefore, \( x eq -2 \).

This restriction subtly limits the set of \( x \) values we can plug into the function. For all \( x \) except \( -2 \), the function \( y = \frac{1}{x + 2} \) is valid and correctly defines \( y \) as a function of \( x \). Understanding domain restrictions helps ensure that computations remain correct and meaningful.