Problem 17 - Beginning and Intermediate Algebra

Question

Solve. $$\frac{2}{x+5}+\frac{1}{x-5}=\frac{16}{x^{2}-25}$$

Step-by-Step Solution

Answer

The short answer to the given equation is \(x=7\).

Step 1: Finding the LCM of the denominators

In order to combine the fractions, we need to find the least common multiple (LCM) of their denominators. The denominators are x+5, x-5, and x^2-25. Observe that \((x^2-25)=(x+5)(x-5)\), so the LCM is simply \(x^2-25\).

Step 2: Combining the fractions

Now, we'll multiply each fraction by the appropriate factor to make all the denominators the same (LCM, which is \(x^2-25\)). This will allow us to combine the fractions: \[\frac{2}{x+5} \cdot \frac{x-5}{x-5} + \frac{1}{x-5} \cdot \frac{x+5}{x+5} = \frac{16}{x^2-25}\] The equation now becomes: \[\frac{2(x - 5) + (x + 5)}{x^2 - 25} = \frac{16}{x^2 - 25}\]

Step 3: Simplifying and solving for x

Since the denominators are the same, we can equate the numerators and solve for x: \[2(x - 5) + (x + 5) = 16\] First, distribute the 2: \[2x - 10 + x + 5 = 16\] Combine the like terms: \[3x - 5 = 16\] Now, isolate x by adding 5 and dividing by 3: \[3x = 21\] \[x = 7\]

Step 4: Checking for excluded values and extraneous solutions

Since the given equation contains fractions, we need to make sure that our solution doesn't cause any of the denominators to become zero. In our equation, x cannot be equal to 5 or -5, as those values will make the denominators equal to zero. Since our solution, x=7, is not an excluded value, it is a valid solution to the given equation. The final answer is \(x=7\).

Key Concepts

Fraction Equations

Fraction equations are mathematical expressions where variables are present in the denominators of fractions. These types of equations can be challenging because they require additional steps to simplify and solve. The goal is usually to find a common denominator so that the fractions can be combined or compared easily.
For example, in the equation \( \frac{2}{x+5}+\frac{1}{x-5}=\frac{16}{x^{2}-25} \), the fractions have different denominators. Addressing these requires identifying a common structure or pattern in the denominators and manipulating the equation accordingly.
This might involve factoring polynomials or using algebraic identities to rewrite the expression in a more convenient form. Solving fraction equations is an essential skill in algebra, helping you to think critically about how numbers and variables interact.

Least Common Multiple

The least common multiple (LCM) is a critical concept when dealing with fraction equations. It is the smallest number that is a multiple of each of the denominators in an equation. Finding the LCM is necessary to make all terms in an equation uniform, which allows you to perform operations like addition and subtraction.
In the given exercise, the denominators are \( x+5 \), \( x-5 \), and \( x^2-25 \). Notice that \( x^2-25 \) factors into \((x+5)(x-5)\), allowing it to be the LCM of the given denominators. This means every fraction can be rewritten with \( x^2-25 \) in the denominator, enabling you to combine the fractions easily.
Recognizing patterns through factoring and simplifying is crucial when determining the LCM and efficiently solving fraction equations.

Extraneous Solutions

Extraneous solutions are potential solutions found during the process of solving equations that turn out not to be valid in the original equation. These occur frequently in fraction equations where variable expressions appear in the denominators.
During solving, operations such as multiplying by the LCM can introduce solutions that make the original denominators zero, which is not allowed since division by zero is undefined.
In our example, the values \( x = 5 \) and \( x = -5 \) would make the denominators zero. Therefore, once a solution is found, it must be checked against these critical values to confirm its validity. This check ensures that an extraneous solution does not slip through the calculations unnoticed.

Polynomial Denominators

Polynomial denominators are expressions containing variables raised to a power, forming a more complex structure than a basic linear denominator. They require special attention when solving fraction equations.
When dealing with polynomial denominators like \( x^2-25 \), one approach is to factor the polynomial to simplify the equation. Factoring allows you to identify common denominators and potential problem spots where division by zero might occur.
For instance, \( x^2-25 \) factors into \( (x+5)(x-5) \), which dictated how we handled the entire fraction equation. Understanding and manipulating polynomial expressions helps streamline solving equations and identifying valid solutions.

• Think of polynomial denominators as puzzle pieces; correctly piecing them together can reveal the pathway to a solution.
• Being familiar with standard algebraic identities and factoring techniques can make this task significantly easier.