Problem 48 - College Algebra

Question

Exercises \(31-50\) contain equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation. $$ \frac{4}{x+5}+\frac{2}{x-5}=\frac{32}{x^{2}-25} $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = \frac{16}{3}\) with restrictions \(x \neq -5, 5\).

Step 1: Identify the Restrictions

First, find the restrictions by setting each denominator equal to zero and solving for \(x\). So, \(x+5=0\) gives \(x=-5\) and \(x-5=0\) gives \(x=5\). The values \(x=-5\) and \(x=5\) will result in a denominator of zero, hence these are the restrictions for the equation. \(x\) can be any real number except -5 and 5.

Step 2: Finding Common Denominator

Next step is to identify a common denominator which in this case is \(x^{2}-25\). Using difference of squares, \(x^{2}-25\) can be factorized as \((x-5)(x+5)\). Multiply each fraction in the equation by \((x-5)(x+5)\) to remove the denominators.

Step 3: Solve the Equation

After removing the denominators, the equation becomes \(4(x-5) + 2(x+5) - 32 = 0\). Solve this equation by combining like terms and simplifying. Doing so results in \(6x - 32 = 0\). Solving for \(x\), we find that \(x = \frac{32}{6} = \frac{16}{3}\).

Step 4: Checking the Answer

Finally, check if the value of \(x = \frac{16}{3}\) violates our restrictions. Since -5 and 5 are our restrictions, \(x = \frac{16}{3}\) is an acceptable solution because it does not equal to these restriction values.

Key Concepts

Restrictions on Variables

When solving rational equations, it's crucial to identify any restrictions on the variables involved. These restrictions occur when a variable makes the denominator of any fraction equal to zero. This is problematic because division by zero is undefined. To find these restrictions, set each denominator equal to zero and solve for the variable.

• For the equation \( \frac{4}{x+5}+\frac{2}{x-5}=\frac{32}{x^{2}-25} \), we set \( x+5=0 \) and \( x-5=0 \) to find restrictions.
• Solving these gives \( x = -5 \) and \( x = 5 \), so these are the values that restrict \( x \).

Therefore, \( x \) can be any real number except -5 and 5, as these would make one or more denominators zero. Always keep these restrictions in mind as they dictate which solutions are valid.

Common Denominator

A common denominator is essential for solving rational equations because it allows you to eliminate the fractions and work with whole numbers instead.

• In our exercise, the original denominators are \( x+5 \), \( x-5 \), and \( x^2-25 \).
• The expression \( x^2-25 \) is a 'difference of squares,' which can be rewritten as \( (x-5)(x+5) \).
• Thus, the least common denominator for all fractions is \( (x-5)(x+5) \).

Multiply each term in the equation by this expression to clear the fractions, making it much easier to solve.

Solving Equations

After determining the common denominator and rewriting the equation, you're ready to solve it. This process starts with the following step: multiply each part of the equation by the common denominator, \( (x-5)(x+5) \). This yields an equation without fractions.

• The equation becomes: \( 4(x-5) + 2(x+5) - 32 = 0 \).
• Distribute and simplify by combining like terms, resulting in \( 6x - 32 = 0 \).

To isolate \( x \), add 32 to both sides, then divide by 6 to find \( x = \frac{16}{3} \). Always check if the solution violates any restrictions identified earlier.

Difference of Squares

The difference of squares is a special type of algebraic expression that can be a key step in solving rational equations. It takes the form \( a^2 - b^2 \) and can be factored as \( (a-b)(a+b) \). This property enables us to factor complex denominators efficiently, as seen in our exercise.

• In this case, \( x^2-25 \) is the difference of squares, equaling \( (x-5)(x+5) \).
• This factorization helps identify the common denominator and simplifies multiplying terms easier.

Understanding and recognizing difference of squares quickly can simplify complex equations and speed up the process of finding a solution.