Problem 23 - Precalculus 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 \(xy + 2y = 1\) defines y as a function of x. After isolating y, we get \(y = \frac{1}{x+2}\), which shows that for each x in the domain, there's exactly one y in the co-domain.

Step 1: Setting the given equation

The equation provided in the exercise is \(xy+2y=1\).

Step 2: Isolate y

To determine if y is a function of x, we should isolate y. We can do this by first factoring out y from the left side of the equation, we get \(y(x+2)=1\).

Step 3: Solve for y

Next, to completely isolate y, we divide each side of the equation by \(x + 2\) to get \(y = \frac{1}{x+2}\).

Key Concepts

Defining Functions of x

Understanding what it means for an equation to define 'y' as a function of 'x' is fundamental in precalculus. A function of 'x' implies that for each input 'x', there is exactly one output 'y'. This unique relationship means that if you were to substitute any value for 'x', there should be only one corresponding 'y' value.
In the context of the given exercise, the question is whether 'y' can be expressed solely in terms of 'x' without ambiguity. When the equation is manipulated to isolate 'y'—as shown in the solution steps—the result is an expression where 'y' depends uniquely on 'x'. The expression \(y = \frac{1}{x+2}\) clearly demonstrates that for any given 'x', there is one specific 'y' value, thus 'y' is a function of 'x'. It's important to note that certain 'x' values could be excluded, such as those that lead to division by zero, which would disrupt the function's definition.

Isolating Variables

The process of isolating variables, like 'y' in our example, is a critical skill in algebra and precalculus. It involves rearranging an equation so that the variable of interest stands alone on one side of the equation. The goal is to have the variable, such as 'y', expressed as explicitly as possible in relation to the other variables or constants.
To isolate 'y', you often perform operations that 'undo' the equation's complexity. For instance, in the exercise, 'y' was initially tangled up in terms with 'x'. The solution began by factoring 'y' out from the terms 'xy + 2y', resulting in \(y(x+2) = 1\). Here, 'y' was not totally isolated, as it was still being multiplied by \(x+2\). The next step was dividing both sides of the equation by \(x+2\) to completely untangle 'y'. The final result, \(y = \frac{1}{x+2}\), offers a clear, isolated view of 'y' in relationship to 'x'. This is pivotal because it allows for an understanding of how 'y' changes as 'x' changes.

Factoring in Algebra

Factoring is a powerful tool in algebra that simplifies expressions and solves equations. It involves breaking down an expression into products of simpler factors that, when multiplied together, result in the original expression. It can transform complicated terms into forms that are easier to manipulate and understand, especially when isolating variables or finding roots.
In our exercise, we see factoring in action when 'y' is factored out from the terms 'xy + 2y', producing the factored form \(y(x+2)\). This step is crucial because it simplifies the left side of the equation to a single product involving 'y'. After factoring, you are often just a step away from isolating the variable, in this case, by dividing both sides by the remaining factor \(x+2\). It’s always beneficial to be comfortable with factoring, as it frequently paves the way for solving complex problems in a more manageable way.