Let's solve the equation step by step:

The given equation is:
\[
-\frac{2}{3} + 2g = -\frac{3}{2}g + 3 - 2g
\]

### Step 1: Combine like terms
#### On the left-hand side (LHS):
\[
-\frac{2}{3} + 2g
\]

#### On the right-hand side (RHS):
\[
-\frac{3}{2}g + 3 - 2g
\]

Simplify the RHS:
\[
-\frac{3}{2}g - 2g + 3 = -\frac{3}{2}g - \frac{4}{2}g + 3 = -\frac{7}{2}g + 3
\]

The equation now becomes:
\[
-\frac{2}{3} + 2g = -\frac{7}{2}g + 3
\]

### Step 2: Move all terms involving \( g \) to one side and constants to the other
Subtract \( 2g \) from both sides:
\[
-\frac{2}{3} = -\frac{7}{2}g - 2g + 3
\]

Simplify the \( g \)-terms on the RHS:
\[
-\frac{7}{2}g - 2g = -\frac{7}{2}g - \frac{4}{2}g = -\frac{11}{2}g
\]

Now the equation is:
\[
-\frac{2}{3} = -\frac{11}{2}g + 3
\]

Next, subtract 3 from both sides:
\[
-\frac{2}{3} - 3 = -\frac{11}{2}g
\]

Simplify the constants on the LHS:
\[
-\frac{2}{3} - 3 = -\frac{2}{3} - \frac{9}{3} = -\frac{11}{3}
\]

The equation now becomes:
\[
-\frac{11}{3} = -\frac{11}{2}g
\]

### Step 3: Solve for \( g \)
Divide both sides by \( -\frac{11}{2} \) (or multiply by the reciprocal):
\[
g = \frac{-\frac{11}{3}}{-\frac{11}{2}} = \frac{11}{3} \cdot \frac{2}{11}
\]

Simplify:
\[
g = \frac{2}{3}
\]

### Final Answer:
\[
g = \frac{2}{3}
\]

---

Would you like more details or have any specific questions? 😊

### Five Related Questions:
1. How do you solve equations involving fractions systematically?
2. Can you explain how to simplify terms with variables and constants?
3. What is the process for finding the least common denominator (LCD) in equations?
4. How do you check if a solution is correct for a given equation?
5. Can such equations be solved graphically, and if so, how?

### Tip:
Always simplify fractions as much as possible early in the solution process to avoid unnecessary complexity.

Learn how to solve the linear equation 2/3 + 2g = -3/2 g + 3 - 2g with a detailed step-by-step solution. This process covers combining like terms, simplifying fractions, and isolating the variable g. Suitable for students in Grades 7-9.

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