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The problem in Question 18 asks to solve the equation \( \frac{x}{2} + \frac{y}{3} = 1 \) for \( x \).

Let's solve this step by step:

1. Start with the equation:
   \[
   \frac{x}{2} + \frac{y}{3} = 1
   \]

2. To isolate \( x \), subtract \( \frac{y}{3} \) from both sides:
   \[
   \frac{x}{2} = 1 - \frac{y}{3}
   \]

3. Now, eliminate the fraction on the left-hand side by multiplying the entire equation by 2:
   \[
   x = 2 \left( 1 - \frac{y}{3} \right)
   \]

4. Simplify the expression:
   \[
   x = 2 - \frac{2y}{3}
   \]

So, the final expression for \( x \) is:
\[
x = 2 - \frac{2}{3}y
\]

Now, comparing this with the provided multiple-choice options:

- The first option is \( x = \frac{2}{3}y \pm 3 \) (not a match).
- The second option is \( x = 3y - 2 \) (not a match).
- The third option is \( x = -\frac{2}{3}y + 3 \) (this is the closest, except for the constant term, which should be \( 2 \) instead of \( 3 \)).

It seems like none of the provided options is an exact match for \( x = 2 - \frac{2}{3}y \), but option 3 is the closest with a slight difference in the constant.

Do you have any questions about this solution, or would you like to see more details?

Here are 5 related questions you might consider:
1. How would you solve for \( y \) instead of \( x \) in the equation \( \frac{x}{2} + \frac{y}{3} = 1 \)?
2. What happens if you substitute specific values for \( y \) and solve for \( x \)?
3. Can you rearrange the equation to express \( y \) as a function of \( x \)?
4. What is the graphical interpretation of the equation \( \frac{x}{2} + \frac{y}{3} = 1 \)?
5. How can you verify the solution by substituting back into the original equation?

**Tip**: Always double-check constants in multiple-choice answers, as they may differ by a small amount from your exact result.

Solve \( \frac{x}{2} + \frac{y}{3} = 1 \) for \( x \).

Learn how to solve the linear equation \( \frac{x}{2} + \frac{y}{3} = 1 \) for x with a detailed step-by-step solution. This problem covers key algebraic concepts like fraction simplification and rearranging linear equations. Suitable for students in Grades 8-10.

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