Problem 88 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(\frac{1}{y+5}=\frac{1}{3 y+6}-\frac{y+2}{y^{2}+7 y+10}\)

Step-by-Step Solution

Answer

Multiply the entire equation by \((y+5)(y+2)\), simplify to find \(y = -5, -2\). Both values make denominators zero, hence invalid solutions.

Step 1: Simplify Denominators

First, let's simplify the denominator \(y^2 + 7y + 10\). Notice that it can be factored: \(y^2 + 7y + 10 = (y+5)(y+2)\). This factorization will help us handle the equation more easily.

Step 2: Rewrite the Equation

Using the factored form of the denominator, rewrite the equation: \(\frac{1}{y+5} = \frac{1}{3(y+2)} - \frac{y+2}{(y+5)(y+2)}\). Now, rewrite \(\frac{1}{3(y+2)}\) as \(\frac{1}{3y+6}\) for consistency.

Step 3: Obtain a Common Denominator

The common denominator for all terms is \((y+5)(y+2)\). Multiply each term by this common denominator to eliminate the fractions: \((y+2) = \frac{1}{(y+5)} \cdot (y+5)(y+2) = (y+2)\), and the right side is \( \cdot \left(\frac{(y+5)(y+2)}{3(y+2)} - (y+2)\right) = \frac{1}{3} (y+5) - (y+2)\).

Step 4: Simplify the Equation

Simplify the obtained equation: \((y+2) = \frac{1}{3} (y+5) - (y+2)\). Simplify by combining like terms: \(-y = \frac{1}{3} (y+5)\), leaving \(-3y = y + 5\).

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial skill in simplifying rational equations. It involves writing a polynomial as a product of its factors, which are simpler polynomials. This process helps in breaking down complex equations and finding solutions effectively. For example, in the expression \( y^2 + 7y + 10 \), we are looking for two numbers that multiply to 10 and add to 7. These numbers are 5 and 2, allowing us to factor the expression as \((y + 5)(y + 2)\).

Factoring makes it easier to identify common denominators in rational expressions and simplifies the equation-solving process by reducing the complexity of fractions. Remember, always check if a polynomial can be factored since it simplifies manipulation across steps in solving equations.

Common Denominators

To solve rational equations efficiently, finding a common denominator is essential. The common denominator allows for all terms in the equation to be combined or compared easily. Here’s a straightforward approach to finding common denominators:

• First, factor the denominators of all fractions involved in the equation.
• Then determine the least common multiple (LCM) of these denominators.
• Rewrite each fraction with this common denominator.

In our example, the expression \(\frac{1}{y+5}=\frac{1}{3(y+2)}-\frac{y+2}{(y+5)(y+2)}\) involves denominators \(y+5\), \(3(y+2)\), and \((y+5)(y+2)\). The common denominator among these is \((y+5)(y+2)\). By multiplying each term by this common denominator, we clear the fractions, allowing us to solve the equation efficiently.

Extraneous Solutions

When solving rational equations, extraneous solutions are those that do not satisfy the original equation, even if they appear valid algebraically. These often arise from steps like multiplying through by the denominator, which can introduce solutions that make the denominator zero, leading to undefined expressions.

Here's how you can handle extraneous solutions:

• After solving the equation, substitute each solution back into the original equation to check for validity.
• If a substituted solution causes any denominator in the original equation to be zero, it is extraneous.
• Only consider solutions that pass this test as valid.

Thoroughly examining potential solutions ensures accuracy and prevents false results in your solution set.