Problem 24 - Algebra for College Students

Question

Find the solution set to each equation. $$\frac{x-4}{x^{2}+2 x-15}=2-\frac{2}{x-3}$$

Step-by-Step Solution

Answer

The solution set is \( x = -13 \).

Step 1: Simplify the equation

First, rewrite the given equation: \[ \frac{x-4}{x^{2}+2x-15}=2-\frac{2}{x-3} \] We will find a common denominator to combine the fractions on the right-hand side. Notice that the left side can be factored: \[ x^2 + 2x - 15 \] factors to \[ (x+5)(x-3) \].

Step 2: Rewrite the right-hand side with a common denominator

The common denominator of 2 and \( \frac{2}{x-3} \) is \( x-3 \). Rewrite both fractions with this common denominator: \[ 2 - \frac{2}{x-3} = \frac{2(x-3)-2}{x-3}= \frac{2x-6-2}{x-3} = \frac{2x-8}{x-3} \] which simplifies to \[ \frac{2(x-4)}{x-3} \].

Step 3: Set the numerators equal

Now, we have: \[ \frac{x-4}{(x+5)(x-3)} = \frac{2(x-4)}{x-3} \]. Since the denominators are different, cross multiply to set the numerators equal: \[ (x-4)(x-3) = 2(x-4)(x+5) \].

Step 4: Solve the equation

Remove the common factor \( x-4 \) from both sides (note: \( x eq 4 \)): \[ x-3 = 2(x+5) \]. Simplify and solve for x: \[ x-3 = 2x + 10 \rightarrow -3 - 10 = x \rightarrow -13 = x. \]

Step 5: Verify the solution

Check the solution in the original equation to ensure it does not make the denominator zero and that it satisfies the equation. For \( x = -13 \): \[ \frac{-13-4}{(-13)^2+2(-13)-15} = \frac{-17}{169 - 26 - 15} = \frac{-17}{128} \]. Check the right-hand side: \[ 2 - \frac{2}{-16} = 2 + \frac{1}{8} = \frac{-17}{128}. \] The solution \( x = -13 \) works.

Key Concepts

Factoring Polynomials

Factoring polynomials is a key skill when solving rational equations. It involves breaking down a complex polynomial into simpler polynomials that multiply together to give the original polynomial. For example, consider the polynomial equation from the exercise: \[ x^2 + 2x - 15 \]. To factor this, we look for two numbers that multiply to -15 and add to 2. These numbers are 5 and -3. So, we can factor the polynomial as: \[ (x+5)(x-3) \]. Factoring helps to simplify the equation and makes it easier to find common denominators or solve for the variable.

Common Denominator

Finding a common denominator is essential when dealing with rational equations, especially when adding or subtracting fractions. In the given exercise, we need a common denominator for \[ 2 - \frac{2}{x-3} \]. The common denominator here is \[ x-3 \]. This allows us to rewrite the problem and simplify it. For example, \[ 2 \] can be rewritten as \[ \frac{2(x-3)}{x-3} \]. Bringing it all together, we get: \[ 2 - \frac{2}{x-3} = \frac{2(x-3) - 2}{x-3} = \frac{2x-8}{x-3} \]. Using a common denominator simplifies the rational equation, making it easier to solve.

Cross Multiplying

Cross multiplying is a method used to solve equations involving fractions. The concept is straightforward: When you have a proportion \[ \frac{a}{b} = \frac{c}{d} \], you can cross multiply to get \[ a \cdot d = b \cdot c \]. In the exercise, after simplifying the equation, we have: \[ \frac{x-4}{(x+5)(x-3)} = \frac{2(x-4)}{x-3} \]. To solve it, we cross multiply: \[ (x-4)(x-3) = 2(x-4)(x+5) \]. This step removes the fractions and simplifies to \[ x-3 = 2(x+5) \], since the common factor \[ x-4 \] can be canceled out from both sides, provided \[ x eq 4 \] to avoid division by zero.

Verifying Solutions

Verifying solutions ensures the answers satisfy the original equation and do not create undefined expressions. To verify, substitute the solution back into the original equation. For \[ x = -13 \], we substitute into \[ \frac{x-4}{x^2+2x-15} = 2 - \frac{2}{x-3} \]. The left side becomes: \[ \frac{-13-4}{(-13)^2 + 2(-13) - 15} = \frac{-17}{169 - 26 - 15} = \frac{-17}{128} \]. The right side: \[ 2 - \frac{2}{-16} = 2 + \frac{1}{8} \]. When simplified both sides are equal \[ \frac{-17}{128} \]. So, the solution \[ x = -13 \] works. Always check solutions as the last step to avoid extraneous solutions that do not satisfy the original equation.