To determine if \( x = \frac{2}{3} \) is a solution to the quadratic equation \( 9x^2 + 3x - 5 = 6 \), we need to substitute \( x = \frac{2}{3} \) into the left-hand side of the equation and check if it equals the right-hand side.
First, substitute \( x = \frac{2}{3} \) into the left side of the equation:
\[ 9\left(\frac{2}{3}\right)^2 + 3\left(\frac{2}{3}\right) - 5 \]
Calculating \( \left(\frac{2}{3}\right)^2 \):
\[ \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]
Now substitute this back in:
\[ 9 \cdot \frac{4}{9} + 3 \cdot \frac{2}{3} - 5 \]
This simplifies to:
\[ 4 + 2 - 5 = 1 \]
Now, the left-hand side equals 1, and the right-hand side is 6:
\[ 1 \neq 6 \]
Since the left side does not equal the right side, \( x = \frac{2}{3} \) is not a solution to the equation.
The correct reasoning is: No, it is not a solution because substituting it back into the equation results in \( 1 = 6 \), which is a false statement.