Problem 15 - College Algebra

Question

For the following exercises, determine whether the relation represents \(y\) as a function of \(x\). $$ x=\frac{3 y+5}{7 y-1} $$

Step-by-Step Solution

Answer

Yes, \(y\) is a function of \(x\), except for \(x = \frac{3}{7}\).

Step 1: Check Relation Structure

The exercise asks whether we can express the relation in the form of a function, where for each value of \(x\) there is exactly one value of \(y\). In this case, we have the relation \(x = \frac{3y + 5}{7y - 1}\). This is a rational expression relating \(x\) to \(y\).

Step 2: Analyze the Equation for One-to-One Mapping

To determine if \(y\) can be written as a function of \(x\), rearrange the equation to solve for \(y\) in terms of \(x\). Begin by cross-multiplying: \(x(7y - 1) = 3y + 5\). Expanding this, you get \(7xy - x = 3y + 5\).

Step 3: Isolate Terms Involving y

To isolate terms involving \(y\), rearrange to get \(7xy - 3y = x + 5\). Factor out \(y\) from the left-hand side to yield \(y(7x - 3) = x + 5\).

Step 4: Solve for y

Now solve for \(y\) by dividing both sides by \((7x - 3)\): \(y = \frac{x + 5}{7x - 3}\). This expression shows \(y\) as a function of \(x\), written as \(y = f(x)\), because for each \(x\) value, there is one corresponding \(y\) value as long as \(7x - 3 eq 0\).

Step 5: Check for Function Validity

To ensure \(y\) remains a function of \(x\), check the restriction \(7x - 3 eq 0\). Solving for \(x\), we have \(7x eq 3\), meaning \(x eq \frac{3}{7}\). Since this is the only restriction, everywhere else \(y\) is a valid function of \(x\).

Key Concepts

Rational Expressions

Rational expressions are a key component of algebra involving fractions where the numerator and/or the denominator are polynomials. In our given problem, the rational expression \(x = \frac{3y + 5}{7y - 1}\) represents a relation between \(x\) and \(y\). To work with rational expressions, it’s crucial to understand how to manipulate and simplify these expressions.

Here are some fundamental points:

• A rational expression is undefined for any value of \(y\) that makes the denominator zero. In our expression, \(7y - 1 = 0\) leads to division by zero, meaning those \(y\) values are not allowed.
• The simplifying process often involves factoring polynomials and reducing the expression by canceling common factors in the numerator and denominator.
• Knowing the restrictions on variables is key, as these will affect the validity of the expression in question.

In this exercise, ensuring that \(7y - 1 eq 0\) keeps the expression valid. Hence, for any other values, the relation continues to define \(x\) in terms of \(y\).

Solving Equations

Solving equations involves techniques for finding the unknown variable. When given a rational expression, you might need to solve for one variable by isolating it on one side of the equation. This is essential in determining one-to-one mappings.

Let's break down the process in this exercise:

• Start with the expression \(x = \frac{3y + 5}{7y - 1}\). To make progress, cross-multiply to eliminate the fraction. This rewrites our equation as \(x(7y - 1) = 3y + 5\).
• Next, distribute \(x\) within so you have \(7xy - x = 3y + 5\).
• To isolate terms involving \(y\), combine similar terms leading to the setup \(7xy - 3y = x + 5\).
• The purpose here is to factor \(y\) to get it by itself. So, factoring results in \(y(7x - 3) = x + 5\).
• Finally, solve for \(y\) to express it as a function of \(x\) with \(y = \frac{x + 5}{7x - 3}\).

Each step involves critical algebraic manipulations to successfully solve the equation.

Function Validity

Function validity checks whether a relation can consistently define one variable as a function of another. This validity comes from ensuring each input \(x\) is linked to only one output \(y\). In a valid function, no \(x\) can have multiple \(y\) values.

In our problem, we derived \(y = \frac{x + 5}{7x - 3}\), a formula intended to express \(y\) as a function of \(x\).

• The validity condition here is \(7x - 3 eq 0\). Solving \(7x = 3\) gives \(x = \frac{3}{7}\), a point where the function is undefined due to division by zero.
• Therefore, \(x eq \frac{3}{7}\) is critical for ensuring every \(x\) has a single \(y\), maintaining the integrity of the function.
• Besides such singular points, rational expressions like \(y = \frac{x + 5}{7x - 3}\) behave as valid functions for other \(x\) values.

This concept emphasizes that a function needs both a consistent output for each input and careful attention to excluded values where the function is undefined.