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Chapter 7: Problem 41

Solve. If no solution exists, state this. $$\frac{3}{x-3}+\frac{5}{x+2}=\frac{5 x}{x^{2}-x-6}$$

Short Answer

Expert verified
No solution exists as \( x = 3 \) makes the denominators zero.

Step by step solution

01

- Identify the denominators

The denominators in the equation are \(x-3\), \(x+2\), and \(x^2-x-6\). The denominator \(x^2-x-6\) can be factored.
02

- Factor the quadratic denominator

Factor \(x^2-x-6\) to get \(x^2-x-6=(x-3)(x+2)\). This means the common denominator for the equation is \( (x-3)(x+2) \)
03

- Rewrite the equation with the common denominator

Rewrite the entire equation using the common denominator \((x-3)(x+2)\): \[ \frac{3(x+2)}{(x-3)(x+2)} + \frac{5(x-3)}{(x-3)(x+2)} = \frac{5x}{(x-3)(x+2)} \]
04

- Combine the fractions

Combine the fractions on both sides of the equation: \[ \frac{3(x+2) + 5(x-3)}{(x-3)(x+2)} = \frac{5x}{(x-3)(x+2)} \]
05

- Simplify the numerator

Expand and simplify the numerator: \[ \frac{3x+6+5x-15}{(x-3)(x+2)} = \frac{5x}{(x-3)(x+2)} \Rightarrow \frac{8x-9}{(x-3)(x+2)} = \frac{5x}{(x-3)(x+2)} \]
06

- Solve the equation

Since the denominators are the same, equate the numerators: \[ 8x - 9 = 5x \]. Solve for \(x\): \[ 8x - 5x - 9 = 0 \Rightarrow 3x - 9 = 0 \Rightarrow x = 3 \]
07

- Check for extraneous solutions

Substitute \(x = 3\) back into the original denominators to ensure the solution doesn't make any denominators zero: The denominators \((x-3)\) and \((x^2 - x - 6)\) become zero, making \(x = 3\) an invalid solution. Therefore, no solution exists.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Quadratics
Quadratic expressions are commonly written in the form of \(ax^2 + bx + c\). To factor a quadratic, we look for two binomials that multiply together to give the original quadratic expression. For example, the expression \(x^2 - x - 6\) can be factored into \((x - 3)(x + 2)\). This is because when we expand \((x - 3)(x + 2)\), we get back the original quadratic: \begin{align*} (x - 3)(x + 2) &= x(x) + x(2) - 3(x) - 3(2) \ &= x^2 + 2x - 3x - 6 \ &= x^2 - x - 6 \right. Factoring helps us simplify rational equations and is crucial for finding common denominators.
Common Denominator
When solving rational equations, having a common denominator is essential for combining and simplifying fractions. In the given problem, the individual denominators \(x-3\), \(x+2\), and \(x^2 - x - 6\) appear in the equation. By factoring the quadratic \(x^2 - x - 6\) to \((x - 3)(x + 2)\), we see that the common denominator is \((x-3)(x+2)\). To rewrite the equation with this common denominator, we multiply: \begin{align*} \frac{3(x+2)}{(x-3)(x+2)} + \frac{5(x-3)}{(x-3)(x+2)} = \frac{5x}{(x-3)(x+2)}. \right. By ensuring all parts of the equation share the same denominator, it becomes easier to combine and compare fractions.
Extraneous Solutions
Extraneous solutions are solutions derived during the process of solving an equation that are not valid in the context of the original problem. These usually emerge when solving rational equations after multiplying through by the common denominator. You must test potential solutions back in the original equation to ensure they don't invalidate any denominators. In the given problem, solving 8x - 9 = 5x yields x = 3. However, substituting x = 3into the denominators x-3 and x^2 - x - 6 makes them zero, indicating division by zero. Hence, x = 3 is extraneous, meaning it is not a valid solution. Therefore, there is no valid solution to the problem.

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