Solve for x, give your answer as a reduced fraction if necessary:

- 7/3 x-1/6=5/9 x - 2/3
  

Let's solve the equation step by step:

\[
\frac{7}{3}x - \frac{1}{6} = \frac{5}{9}x - \frac{2}{3}
\]

### Step 1: Eliminate the fractions by finding a common denominator.
The common denominator of 3, 6, 9 is 18. Multiply every term of the equation by 18 to eliminate the denominators:

\[
18 \times \left( \frac{7}{3}x - \frac{1}{6} \right) = 18 \times \left( \frac{5}{9}x - \frac{2}{3} \right)
\]

This gives:

\[
6 \times 7x - 3 = 2 \times 5x - 6 \times 2
\]

Simplify:

\[
42x - 3 = 10x - 12
\]

### Step 2: Move all terms involving \(x\) to one side and constant terms to the other side.

Subtract \(10x\) from both sides:

\[
42x - 10x - 3 = -12
\]

This simplifies to:

\[
32x - 3 = -12
\]

### Step 3: Isolate the \(x\)-term.

Add 3 to both sides:

\[
32x = -12 + 3
\]

\[
32x = -9
\]

### Step 4: Solve for \(x\).

Divide both sides by 32:

\[
x = \frac{-9}{32}
\]

### Final Answer:
\[
x = \frac{-9}{32}
\]

Would you like further details or clarifications on this solution?

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Here are 5 related questions to expand your understanding:
1. How would the solution change if the coefficients were decimals instead of fractions?
2. How can you verify the solution by substituting \(x = \frac{-9}{32}\) back into the original equation?
3. What if one side of the equation had been quadratic in terms of \(x\)?
4. How would this problem change if there were multiple variables involved?
5. How can you apply this technique to solve similar equations with mixed fractions?

**Tip:** Always look for the least common denominator to quickly eliminate fractions in an equation.

Solve for x, give your answer as a reduced fraction if necessary: 7/3 x - 1/6 = 5/9 x - 2/3

This detailed guide shows how to solve the linear equation 7/3x - 1/6 = 5/9x - 2/3 step by step. Learn how to eliminate fractions by finding a common denominator, isolate variables, and solve for x. Suitable for Grades 8-10, this problem covers key algebra concepts including fractions and linear equations.

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