Problem 39 - Intermediate Algebra

Question

Solve each equation. If a solution is extraneous, so indicate. \(\frac{6}{x+3}+\frac{48}{x^{2}-2 x-15}-\frac{7}{x-5}=0\)

Step-by-Step Solution

Answer

No solution, as the only potential solution \( x = -3 \) is extraneous.

Step 1: Factor the Quadratic

First, let's factor the quadratic in the denominator: \ x^2 - 2x - 15. \ This can be factored into: \ (x - 5)(x + 3). \ Now the equation becomes: \( \frac{6}{x+3} + \frac{48}{(x-5)(x+3)} - \frac{7}{x-5} = 0. \)

Step 2: Find the Least Common Denominator (LCD)

To combine the fractions, we need the least common denominator. The LCD of \(x+3\) and \(x-5\) is \((x+3)(x-5)\).

Step 3: Rewrite Each Fraction with the LCD

\( \frac{6}{x+3} \) is rewritten as \( \frac{6(x-5)}{(x+3)(x-5)} \), \( \frac{48}{(x-5)(x+3)} \) stays the same, and \( \frac{7}{x-5} \) is rewritten as \( \frac{7(x+3)}{(x+5)(x-5)} \). Thus, the equation becomes: \\( \frac{6(x-5) + 48 - 7(x+3)}{(x+3)(x-5)} = 0. \)

Step 4: Simplify the Numerator

Simplify the numerator: Expand the terms: \[ 6(x-5) + 48 - 7(x+3) = 6x - 30 + 48 - 7x - 21. \] Combine like terms: \[ -x - 3. \] Now the equation simplifies to: \( \frac{-x-3}{(x+3)(x-5)} = 0. \)

Step 5: Solve for x

For a fraction to equal zero, its numerator must equal zero. Solve the equation: \( -x - 3 = 0 \). Add 3 to both sides: \( -x = 3 \). Multiply by -1: \( x = -3 \). Also consider denominators: solutions cannot be where denominators are zero: \( x eq 5 \) and \( x eq -3 \). Thus \( x = -3 \) is extraneous.

Step 6: Conclusion

Since the solution \( x = -3 \) is extraneous, the equation has no solution.

Key Concepts

Rational Equations

Rational equations are mathematical statements, involving fractions with polynomials in their numerators and denominators.
Solving these equations requires finding a common denominator. The purpose is to rewrite each term using this common denominator. This allows the fractions to be combined or compared more easily.

• Involves fractions with polynomials.
• Solutions involve finding common denominators.
• Essential for simplifying and solving.

After obtaining the common denominator, the equation simplifies into a form that allows you to solve for the variable. Always consider which values would make the denominator zero. This is crucial because such solutions are not valid in mathematics. Rational equations are a staple of intermediate algebra and knowing how to manipulate them is key to solving many higher-level math problems.

Extraneous Solutions

Extraneous solutions are solutions that emerge when solving equations but aren't valid for the original problem.
These solutions often arise when applying algebraic rules and transformations, especially with equations involving fractions.When you solve rational equations, factor the denominators and identify the values that make them zero.

• These values should be excluded from your solution set from the start.
• If one of your obtained solutions turns out to be one of these values, it becomes extraneous.

For example, in the given problem, solving the simplified equation led to the solution: \( x = -3 \),
which made a denominator zero, hence is extraneous. This illustrates why checking and validating solutions is a vital step.Without doing this, you might mistakenly think you have a valid solution when there is none. Get into the habit of checking if your solution causes any denominators to zero, and you will avoid such mistakes.

Factoring Quadratics

Factoring quadratics involves breaking down a quadratic expression into a product of simpler expressions.
It is one of the most fundamental skills learned in algebra.This process is useful whenever you have a quadratic in an equation, especially when this quadratic is part of a fraction's denominator as seen in rational equations.

• Look for two numbers that multiply to the constant term and add up to the linear coefficient.
• Rewrite the middle term using these two numbers, and factor by grouping.

For instance, the quadratic \(x^2 - 2x - 15\) factors into \((x - 5)(x + 3)\). This not only simplifies complex equations but also helps in identifying restrictions on solutions by finding zeros of the quadratic.
The zeros determine where the original rational expression is undefined. Mastering this skill enhances your ability to manipulate and solve a wide array of algebraic problems efficiently.