Problem 77
Question
Solve each equation. Identify each equation as a conditional equation, an inconsistent equation, or an identity. State the solution sets to the identities using interval notation. $$\frac{1}{x}+\frac{1}{2 x}=\frac{5}{4 x}$$
Step-by-Step Solution
Verified Answer
Inconsistent equation with no solution.
1Step 1: Find a Common Denominator
The common denominator for the terms in the equation \( \frac{1}{x} + \frac{1}{2x} = \frac{5}{4x} \) is \(4x\). Rewrite each term with the common denominator: \( \frac{4}{4x} + \frac{2}{4x} = \frac{5}{4x} \).
2Step 2: Combine the Terms
Combine the fractions on the left side of the equation: \( \frac{4 + 2}{4x} = \frac{5}{4x} \). This simplifies to \( \frac{6}{4x} = \frac{5}{4x} \).
3Step 3: Eliminate the Denominator
Since the denominators are the same, we can eliminate them and focus on the numerators: \(6 = 5\).
4Step 4: Determine the Type of Equation
The equation \(6 = 5\) is not true for any real number, therefore it is an inconsistent equation.
Key Concepts
common denominatorcombining fractionssolving rational equationstypes of equations
common denominator
When working with fractions in algebra, a common denominator is essential to simplify and solve equations. The common denominator is just a shared multiple of the denominators you're working with. Finding it allows you to rewrite fractions so they have the same bottom part, making them easier to combine.
For example, in the problem \(\frac{1}{x} + \frac{1}{2x} = \frac{5}{4x}\), the denominators are 'x,' '2x,' and '4x.' The least common multiple of these values is '4x.' This means all fractions can be rewritten with '4x' as their denominator:
For example, in the problem \(\frac{1}{x} + \frac{1}{2x} = \frac{5}{4x}\), the denominators are 'x,' '2x,' and '4x.' The least common multiple of these values is '4x.' This means all fractions can be rewritten with '4x' as their denominator:
- \(\frac{1}{x} = \frac{4}{4x}\)
- \(\frac{1}{2x} = \frac{2}{4x}\)
- \(\frac{5}{4x}\) is already with the common denominator
- .
With a common denominator, combining and solving becomes straightforward.
combining fractions
After establishing a common denominator, the next step is to combine the fractions. This involves adding the numerators while keeping the common denominator the same.
Consider our rewritten equation: \(\frac{4}{4x} + \frac{2}{4x} \). Here, the denominators are the same ('4x'). So, you only need to add the numerators: \(\frac{4 + 2}{4x} = \frac{6}{4x}\).
Combining fractions in this manner simplifies equations-step by step-making complex problems more manageable. The key is to ensure the denominators are unified, then operate on the numerators.
For clarity:
Consider our rewritten equation: \(\frac{4}{4x} + \frac{2}{4x} \). Here, the denominators are the same ('4x'). So, you only need to add the numerators: \(\frac{4 + 2}{4x} = \frac{6}{4x}\).
Combining fractions in this manner simplifies equations-step by step-making complex problems more manageable. The key is to ensure the denominators are unified, then operate on the numerators.
For clarity:
- Add/Subtract the numerators.
- Retain the common denominator.
- This essentially turns multiple fractions into a single, simplified fraction.
solving rational equations
Rational equations involve fractions with variables in the denominators. Solving them often requires a systematic approach. First, find a common denominator to unify the fractions. Then, combine them as needed.
For instance, in \(\frac{6}{4x} = \frac{5}{4x}\), first confirm the denominators are common. With denominators matched, eliminate them to focus on numerators: \(6 = 5\). However, this isn't always straightforward. The solution depends on the equation type (e.g., conditional, identity, inconsistent).
When denominators cancel naturally, like in our example, it simplifies matters. This process helps in identifying the essence of the equation and deciding if it's solvable.
For instance, in \(\frac{6}{4x} = \frac{5}{4x}\), first confirm the denominators are common. With denominators matched, eliminate them to focus on numerators: \(6 = 5\). However, this isn't always straightforward. The solution depends on the equation type (e.g., conditional, identity, inconsistent).
When denominators cancel naturally, like in our example, it simplifies matters. This process helps in identifying the essence of the equation and deciding if it's solvable.
- Start by finding a common denominator.
- Combine fractions.
- Eliminate the denominator to solve the equation.
types of equations
Understanding different types of equations is vital in math. They include conditional, inconsistent, and identity equations. These distinctions dictate how you'll approach solving them.
A conditional equation has specific solutions that satisfy it. For example, \(2x + 3 = 7\) is conditional, as only \(x = 2\) works. An identity holds true for all values of the variable, like \(x = x\). An inconsistent equation, like \(6 = 5\) (from our example), has no solution since it's always false.
Recognizing these types helps:
A conditional equation has specific solutions that satisfy it. For example, \(2x + 3 = 7\) is conditional, as only \(x = 2\) works. An identity holds true for all values of the variable, like \(x = x\). An inconsistent equation, like \(6 = 5\) (from our example), has no solution since it's always false.
Recognizing these types helps:
- Conditional equations need precise solving.
- Identities require verification.
- Inconsistent equations are unsolvable.
- This knowledge simplifies problem-solving, guiding your approach based on equation type.
Other exercises in this chapter
Problem 76
Perform the indicated operations. $$\frac{(r t)^{3}}{r t^{4}} \div \frac{\left(r t^{2}\right)^{3}}{r^{2} t^{3}}$$
View solution Problem 76
In place of each question mark in Exercises \(75-92,\) put an expression that will make the rational expressions equivalent. $$4=\frac{?}{3}$$
View solution Problem 77
Perform the indicated operations. $$\frac{2 x^{2}+7 x-15}{4 x^{2}-100} \cdot \frac{2 x^{2}-9 x-5}{4 x^{2}-1}$$
View solution Problem 77
In place of each question mark in Exercises \(75-92,\) put an expression that will make the rational expressions equivalent. $$5=\frac{10}{?}$$
View solution