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Chapter 2: Problem 21

Solve each rational equation for x. State all x-values that are excluded from the solution set. \(\frac{1}{x}=\frac{1}{5}+\frac{3}{2 x}\)

Short Answer

Expert verified
The solution is \(x = -\frac{5}{2}\), excluding \(x = 0\).

Step by step solution

01

Identify the common denominator

In the equation \(\frac{1}{x}=\frac{1}{5}+\frac{3}{2x}\), identify a common denominator that can be used to eliminate the fractions. The denominators in this equation are \(x\), \(5\), and \(2x\). The least common denominator (LCD) for these terms is \(10x\).
02

Multiply both sides by the common denominator

Multiply every term in the equation by the common denominator \(10x\) to eliminate the fractions: \[10x \cdot \frac{1}{x} = 10x \cdot \frac{1}{5} + 10x \cdot \frac{3}{2x}\].
03

Simplify the equation

Simplify each term: \(10 = 2x + 15\). This simplification is done by canceling terms in the denominators and multiplying the results.
04

Solve for x

Now solve the equation \(10 = 2x + 15\) to find \(x\). Subtract 15 from both sides: \(-5 = 2x\). Then divide by 2: \(x = -\frac{5}{2}\).
05

State excluded values

Determine which values of \(x\) are excluded by checking where the denominators in the original equation are zero. The denominators \(x\) and \(2x\) are zero when \(x = 0\). So \(x = 0\) is excluded.

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Key Concepts

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

Understanding the Least Common Denominator
When solving rational equations, finding the least common denominator (LCD) is a crucial step. It helps simplify the process of solving equations that contain fractions. In the equation \(\frac{1}{x} = \frac{1}{5} + \frac{3}{2x}\), the denominators are \(x\), \(5\), and \(2x\). The LCD is the smallest number or expression that each denominator can divide without a remainder.
The process involves determining the largest multiple of each term that can encompass all denominators.
  • In this equation, the LCD is \(10x\). This is because 10 is the smallest multiple that both 5 and 2 divide evenly, and including \(x\) accommodates terms with \(x\) and \(2x\).
Once you know the LCD, you multiply each term of the equation by it to clear the fractions, making it much simpler to solve the equation. The goal is to write an equivalent equation devoid of fractions, thereby streamlining calculations.
Identifying Excluded Values
Excluded values are specific values of \(x\) that make any denominator in the equation zero. These values are crucial because division by zero is undefined in mathematics. Hence, any solution that results in these values must be rejected.Considering the equation \(\frac{1}{x} = \frac{1}{5} + \frac{3}{2x}\):
To find excluded values, check when each denominator equals zero.
  • For the term \(\frac{1}{x}\), \(x = 0\) makes the denominator zero.
  • For the term \(\frac{3}{2x}\), \(2x = 0\) also results in zero when \(x = 0\).
Thus, \(x = 0\) is excluded from the solution set. Always determine these values before solving to ensure accuracy in solutions.
The Process of Fraction Elimination
Fraction elimination involves clearing fractions from the equation, a significant simplification technique in algebra. Once the LCD is determined, each fraction is multiplied by it to remove the denominators from the equation.For the equation \(\frac{1}{x} = \frac{1}{5} + \frac{3}{2x}\), we utilize the LCD, \(10x\), to eliminate fractions.
Here's a detailed step-by-step of how this works:
  • Multiply \(\frac{1}{x}\) by \(10x\): The \(x\) in the denominator is canceled by the \(x\) in the LCD, leaving \(10\).
  • Multiply \(\frac{1}{5}\) by \(10x\): The fraction simplifies to \(2x\) since \(10x \div 5 = 2x\).
  • Multiply \(\frac{3}{2x}\) by \(10x\): The \(2x\) in the denominator cancels partially with \(10x\), simplifying to \(15\).
After these operations, the equation becomes \(10 = 2x + 15\), greatly simplifying the process. This method entirely removes the complexity of handling fractional denominators and makes further steps straightforward.

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Most popular questions from this chapter

For the following exercises, use the model for the period of a pendulum, \(T\), such that \(T=2 \pi \sqrt{\frac{L}{g}}\), where the length of the pendulum is \(L\) and the acceleration due to gravity is \(g\). If the acceleration due to gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) and the period equals \(1 \mathrm{~s}, \mathrm{fi}\) d the length to the nearest \(\mathrm{cm}\) \((100 \mathrm{~cm}=1 \mathrm{~m})\)

For the following exercises, input the left-hand side of the inequality as a Y1 graph in your graphing utility. Enter \(\mathrm{Y} 2=\) the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, liabs ( Find the points of intersection, recall \((2^{\text { nd }}\) CALC 5:intersection, lst curve, enter, } \(2^{\text { nd }}\) curve, enter, guess, enter). Copy a sketch of the graph and shade the \(x\) -axis for your solution set to the inequality. Write final answers in interval notation. $$ |x-4|<3 $$

For the following exercises, solve the compound inequality. Express your answer using inequality signs, and then write your answer using interval notation. $$ 3 y<5-2 y<7+y $$

For the following exercises, describe all the \(x\) -values within or including a distance of the given values. Distance of 11 units from the number 1

For the following exercises, input the left-hand side of the inequality as a Y1 graph in your graphing utility. Enter \(\mathrm{Y} 2=\) the right-hand side. Entering the absolute value of an expression is found in the MATH menu, Num, liabs ( Find the points of intersection, recall \((2^{\text { nd }}\) CALC 5:intersection, lst curve, enter, } \(2^{\text { nd }}\) curve, enter, guess, enter). Copy a sketch of the graph and shade the \(x\) -axis for your solution set to the inequality. Write final answers in interval notation. $$ |x+2| \geq 5 $$
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