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Magazine Chapter 7: Problem 109
Solve. $$\frac{15}{x}-\frac{15}{x+2}=2$$
Short Answer
Expert verified
The solutions are \( x = -5 \) and \( x = 3 \).
Step by step solution
01
- Find a common denominator
To solve the equation \(\frac{15}{x}-\frac{15}{x+2}=2\), first find a common denominator for the fractions. The common denominator for \(x\) and \(x+2\) is \(x(x+2)\).
02
- Rewrite each term
Rewrite each term with the common denominator: \[ \frac{15(x+2)}{x(x+2)} - \frac{15x}{x(x+2)} = 2 \]
03
- Combine the fractions
Combine the fractions over the common denominator:\[ \frac{15(x+2) - 15x}{x(x+2)} = 2 \] This simplifies to: \[ \frac{30}{x(x+2)} = 2 \]
04
- Cross multiply
Cross multiply to clear the fractions:\[ 30 = 2x(x+2) \]
05
- Expand and simplify
Expand the right side and simplify:\[ 30 = 2x^2 + 4x \]Rearrange the equation to standard form:\[ 2x^2 + 4x - 30 = 0 \]
06
- Solve the quadratic equation
Solve the quadratic equation. First, divide everything by 2 for simplicity:\[ x^2 + 2x - 15 = 0 \]Factorize the quadratic:\[ (x + 5)(x - 3) = 0 \]So, the solutions are:\[ x = -5 \] and \[ x = 3 \]
07
- Check the solutions
Check both solutions in the original equation:For \( x = -5 \): \[ \frac{15}{-5} - \frac{15}{-5+2} = -3 - (-5) = -3 + 5 = 2 \] which is true.For \( x = 3 \): \[ \frac{15}{3} - \frac{15}{3+2} = 5 - 3 = 2 \] which is also true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Common Denominators
When solving rational equations like the one given, finding a common denominator is crucial. This simplifies the equation into a single fraction. For example, let's take the equation \( \frac{15}{x} - \frac{15}{x+2} = 2 \). The denominators are \(x\) and \(x+2\). To combine these fractions, determine the common denominator, which is \(x(x+2)\). This common denominator allows us to rewrite each term of the equation, making subsequent steps more manageable. Remember: having a single denominator streamlines the process, paving the way for easier manipulation and solution.
Cross Multiplication
Cross multiplication is a powerful tool when working with rational equations. Once you have a single fraction, like \( \frac{30}{x(x+2)} = 2 \), cross multiplying clears the fractions, converting the equation into a polynomial form. Cross multiplying means you multiply the numerator of one fraction by the denominator of the other. For example, multiplying both sides of \( \frac{30}{x(x+2)} = 2 \) by \(x(x+2)\) gives you: \[ 30 = 2x(x+2) \]. This step eliminates the fraction, allowing you to work with a standard algebraic equation.
Factoring Quadratics
Factoring quadratics is essential when simplifying equations during problem-solving. After cross multiplication and simplification, you may end up with a quadratic equation like \( 2x^2 + 4x - 30 = 0 \). Dividing by 2 simplifies it to \( x^2 + 2x - 15 = 0 \). To factorize, find two numbers that multiply to -15 and add up to 2, which are 5 and -3. Hence, the quadratic becomes: \((x + 5)(x - 3) = 0\). This factorization reveals the possible solutions for \(x\), which are \(x = -5\) and \(x = 3\).
Checking Solutions
Always check your solutions in the original equation to verify their validity. Plugging \(x = -5\) into \( \frac{15}{x} - \frac{15}{x+2} = 2 \) yields: \[ \frac{15}{-5} - \frac{15}{-5+2} = -3 - (-5) = -3 + 5 = 2 \]. Similarly, for \(x = 3\): \[ \frac{15}{3} - \frac{15}{3+2} = 5 - 3 = 2 \]. Since both values satisfy the original equation, they are valid solutions. This final step ensures the integrity of your answers and prevents mistakes from overlooked errors.
