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Chapter 2: Problem 74

For exercises 65-86, (a) solve. (b) check. $$ \frac{2}{3}(n-1)+\frac{1}{6}=n-\frac{1}{2} $$

Short Answer

Expert verified
n = 0

Step by step solution

01

- Distribute the fraction

Distribute \(\frac{2}{3}\) through the parenthesis: \(\frac{2}{3}(n-1)\) becomes \[ \frac{2}{3}n - \frac{2}{3} \].
02

- Combine like terms

Rewrite the equation with the distributed values: \[ \frac{2}{3}n - \frac{2}{3} + \frac{1}{6} = n - \frac{1}{2} \]. Combine like terms on the left side by finding a common denominator (6): \[ \frac{2}{3}n - \frac{4}{6} + \frac{1}{6} = \frac{2}{3}n - \frac{3}{6} = \frac{2}{3}n - \frac{1}{2} \].
03

- Simplify the equation

Simplify the left side: \[ \frac{2}{3}n - \frac{1}{2} = n - \frac{1}{2} \]. The equations on both sides are now the same.
04

- Isolate the variable

Since \( \frac{2}{3}n = n \), we need to isolate \(n\). Subtract \(\frac{2}{3}n\) from both sides: \[ 0 = n - \frac{2}{3}n \].
05

- Solve for n

Combine \(n - \frac{2}{3}n\) to find the value: \[ 0 = \frac{1}{3}n \]. Multiplying both sides by 3, \(n = 0\).
06

- Check the solution

Substitute \(n = 0\) back into the original equation: \[ \frac{2}{3}(0-1) + \frac{1}{6} = 0 - \frac{1}{2} \]. Simplify both sides: \[ \frac{2}{3} \times -1 + \frac{1}{6} = -\frac{2}{3} + \frac{1}{6} = -\frac{4}{6} + \frac{1}{6} = -\frac{3}{6} = -\frac{1}{2} \], matching both sides. Thus, the solution \(n = 0\) is confirmed.

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Key Concepts

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

Distributive Property
The distributive property is essential in algebra as it lets you multiply a single term by each term within parentheses. In our problem, we start with the term \(\frac{2}{3}(n-1)\). We distribute \(\frac{2}{3}\) to both \(n\) and \(-1\): \(\frac{2}{3} \times n\) becomes \(\frac{2}{3}n\) and \(\frac{2}{3} \times -1\) becomes \(-\frac{2}{3}\). This step helps to remove parentheses and simplify the equation.
Combining Like Terms
To make an equation easier to solve, combine like terms. Here, we see the equation \(\frac{2}{3}n - \frac{2}{3} + \frac{1}{6} = n - \frac{1}{2}\). The goal is to simplify. Focus on the left side: find a common denominator for the fractions, which is 6. This changes \(-\frac{2}{3}\) to \(-\frac{4}{6}\), letting us combine it with \(\frac{1}{6}\). Simplifying yields \(\frac{2}{3}n - \frac{1}{2}\). Now, both sides look similar, making the equation simpler to handle.
Isolating the Variable
Next, we need to get the variable (in this case, \(n\)) alone. Our simplified equation is \(\frac{2}{3}n - \frac{1}{2} = n - \frac{1}{2}\). Remove \(\frac{1}{2}\) from each side: we're left with \(\frac{2}{3}n = n\). Subtract \(\frac{2}{3}n\) from both sides, giving us \(0 = n - \frac{2}{3}n\). Combine \(n - \frac{2}{3}n\), resulting in \(0 = \frac{1}{3}n\). Finally, multiply by 3 to isolate \(n\): \(n = 0\).
Checking Solutions in Algebra
Always confirm your solution by substituting it back into the original equation. For \(n = 0\), our equation is \(\frac{2}{3}(0-1) + \frac{1}{6} = 0 - \frac{1}{2}\). This simplifies to \(-\frac{2}{3} + \frac{1}{6}\), which further reduces to \(-\frac{1}{2}\), matching both sides. Therefore, our solution \(n = 0\) is correct. Checking solutions ensures accuracy in your work, allowing you to catch any mistakes before finalizing.

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