Problem 41 - College Algebra and Calculus: An Applied Approach

Question

Solve the equation and check your solution. (Some equations have no solution.) $$ \frac{1}{x-3}+\frac{1}{x+3}=\frac{10}{x^{2}-9} $$

Step-by-Step Solution

Answer

The solution to the given equation is \(x = 5\).

Step 1: Simplify the equation

On the left side, we can see two fractions which have the same denominator when multiplied with each other. On the right side, the term \(x^{2} - 9\) can be rewritten as \((x + 3)(x - 3)\), using the difference of squares formula. This will help to simplify the equation. Thus the equation becomes: \[\frac{1}{x-3}+\frac{1}{x+3}=\frac{10}{(x+3)(x-3)}\]

Step 2: Clear the fractions

To clear the fractions, we multiply every term by the LCD \((x-3)(x+3)\), which results in: \[ (x+3)+(x-3)=10\]

Step 3: Simplify

Simplifying step 2, we get \[ 2x=10 \] This equation can now be easily solved by dividing both sides by 2.

Step 4: Solve for x

When you divide 10 by 2, you get \[ x= 5\]

Step 5: Check the solution

Substitute x = 5 back into the original equation to check if it is a valid solution. Be careful, if the solution causes any denominator to become zero, it's an extraneous solution (which is not possible). Luckily, that does not happen in this case, so x = 5 is indeed the solution

Key Concepts

Difference of Squares

The difference of squares is a useful algebraic identity, given as \[ a^2 - b^2 = (a - b)(a + b) \]. This trick helps to factorize expressions involving two terms squared with a subtraction in between. Understanding this concept can greatly simplify solving rational equations.
This problem involves the term \( x^2 - 9 \), which can be seen as a difference of squares because \( 9 \) is \( 3^2 \).
Thus, we rewrite it as \[ x^2 - 9 = (x - 3)(x + 3) \].

• This transformation allows us to match related denominators across rational terms.
• A simplified version of the expression helps in easily multiplying and dividing terms in the equation.

With this transformation, solving the equation becomes straightforward, since it allows clearing fractions by eliminating the zero denominator potential.

Checking Solutions

When solving equations, especially rational ones, checking solutions is crucial. This ensures that the value we solve for truly satisfies the original equation.
After finding \( x = 5 \), substitute it back into the original equation to confirm it's correct:

• Calculate the left side: \( \frac{1}{5-3} + \frac{1}{5+3} = \frac{1}{2} + \frac{1}{8} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8} \).
• Calculate the right side with \( x = 5 \): \( \frac{10}{5^2 - 9} = \frac{10}{25 - 9} = \frac{10}{16} = \frac{5}{8} \).

If both sides equal, the solution is verified as correct. If they differ, it's back to exploring other potential solutions or re-evaluating the solving steps. Checking prevents errors and confirms our final solution.

Extraneous Solutions

Extraneous solutions often occur with rational equations. While solving, especially when multiplying or assuming certain algebraic manipulations, it might appear as if there's a solution when there isn't.An extraneous solution seems like it works from transformed versions of the equation but fails in the original form due to resulting in zero denominators.

• Always substitute potential solutions back into the original equation to check validity.
• If substituting causes any denominator to be zero, discard it as extraneous.

In our case, \( x = 5 \) does not cause any zero denominators, thus it's a valid solution.
Understanding extraneous solutions prevents false conclusions and ensures accuracy in mathematical problem-solving.