Problem 9 - Beginning and Intermediate Algebra

Question

Solve. $$\frac{y^{2}}{y+4}=\frac{16}{y+4}$$

Step-by-Step Solution

Answer

The short answer to the problem is: \(y = 4\).

Step 1: Equate the Numerators

Since both sides of the equation have the same denominator, we can equate the numerators: \(y^2 = 16\)

Step 2: Solve for \(y\)

Now, to find the values of \(y\), take the square root of both sides: \(\sqrt{y^2} = \pm\sqrt{16}\)

Step 3: Simplify

Simplify the square roots: \(y = \pm 4\)

Step 4: Check for Extraneous Solutions

Since the original equation has a denominator of \(y + 4\), we need to make sure that our solutions don't make the denominator equal to zero. If \(y = -4\), then the denominator would be zero, creating an undefined expression. So, we exclude \(y=-4\) as an extraneous solution. Therefore, the only valid solution is \(y = 4\).

Key Concepts

Numerator and Denominator

In solving rational equations, identifying and working with the numerator and denominator is crucial. These equations involve fractions, where values (variables or numbers) appear above and below a fraction line.

• The **numerator** is the top part of a fraction. It indicates how many parts of the whole (expressed by the denominator) are being considered.
• The **denominator** is the bottom part. It shows how many equal parts the whole is divided into.

In the original exercise, both fractions have the same denominator \(y + 4\). When fractions have the same denominator, it simplifies the process as we can directly compare their numerators. This is because if two fractions are equal and their denominators are the same, then their numerators must also be equal.By equating the numerators, we convert the fraction equation into a simpler quadratic equation \(y^2 = 16\), making it easier to solve.

Extraneous Solutions

Extraneous solutions are solutions derived from the algebraic manipulation of an equation that do not satisfy the original equation. They often occur when both sides of an equation are squared or when dealing with fractional equations.In the given problem, after solving \(y^2 = 16\), we find potential solutions \(y = 4\) and \(y = -4\). However, solutions in rational equations must always be checked to see if they make any denominators in the original equation equal to zero. This is because a zero denominator leads to undefined expressions.

• In our case, we need to check the denominator \(y + 4\).
• Substituting \(y = -4\) into \(y + 4\) gives zero, making the expression undefined.

Hence, \(y = -4\) is an extraneous solution, and must be excluded, leaving \(y = 4\) as the only valid solution.

Square Roots

Understanding square roots is essential when solving equations like \(y^2 = 16\). A square root asks 'what number can be multiplied by itself to give the original number?'

• The square root of 16 can be both 4 and -4 since \(4 \times 4 = 16\) and \((-4) \times (-4) = 16\).
• This is why when solving \(y^2 = 16\), we get two potential solutions: \(y = 4\) and \(y = -4\).

Always remember, when taking the square root of both sides of an equation, include both the positive and negative roots. In the context of rational equations, like in our problem, verifying these solutions within the original equation is key to identifying any extraneous roots. This ensures accurate and meaningful solutions, contributing to a deeper understanding of how square roots work in algebra.