Prove that if the equation `x^2+9y^2-4x+3=0` is satisfied for real values of `xa n dy ,t h e nx` must lie between 1 and 3 and`y` must lie between-1/3 and 1/3.

Explanation

To prove that if the equation \( x^2 + 9y^2 - 4x + 3 = 0 \) is satisfied for real values of \( x \) and \( y \), then \( x \) must lie between 1 and 3, and \( y \) must lie between \(-\frac{1}{3}\) and \(\frac{1}{3}\), we can follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ x^2 + 9y^2 - 4x + 3 = 0 \]
Rearranging gives:
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