Let's solve the equation:

\[
x + \frac{1}{6} = 4 - 2x
\]

### Step 1: Move all terms involving \(x\) to one side
To do this, add \(2x\) to both sides:

\[
x + 2x + \frac{1}{6} = 4
\]

This simplifies to:

\[
3x + \frac{1}{6} = 4
\]

### Step 2: Subtract \(\frac{1}{6}\) from both sides
\[
3x = 4 - \frac{1}{6}
\]

### Step 3: Simplify the right-hand side
We can express \(4\) as a fraction with denominator 6:

\[
4 = \frac{24}{6}
\]

Now subtract:

\[
3x = \frac{24}{6} - \frac{1}{6} = \frac{23}{6}
\]

### Step 4: Solve for \(x\)
To isolate \(x\), divide both sides by 3:

\[
x = \frac{23}{6} \times \frac{1}{3} = \frac{23}{18}
\]

Thus, the solution is:

\[
x = \frac{23}{18}
\]

Would you like more details on any step?

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Here are 5 related questions to further explore:

1. How do you simplify an equation with fractions on both sides?
2. What happens if we encounter equations with variables in the denominator?
3. How would you solve a quadratic equation like \(x^2 + 3x = 2\)?
4. How can you solve an equation with decimals instead of fractions?
5. How do you handle equations involving absolute values?

**Tip:** Always try to clear fractions early in an equation by multiplying through by the least common denominator (LCD).

This page provides a detailed solution to the linear equation x + 1/6 = 4 - 2x, including steps to simplify and solve equations involving fractions. Learn algebraic techniques suitable for Grades 6-8 students, with clear explanations of each step.

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