To determine if \( x = \frac{2}{3} \) is a solution to the equation \( 9x^2 + 3x - 5 = 6 \), we can substitute \( \frac{2}{3} \) into the left side of the equation and see if it equals 6.
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Substitute \( x = \frac{2}{3} \) into the left side of the equation: \[ 9\left(\frac{2}{3}\right)^2 + 3\left(\frac{2}{3}\right) - 5 \]
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Calculate \( 9\left(\frac{2}{3}\right)^2 \): \[ \left(\frac{2}{3}\right)^2 = \frac{4}{9} \implies 9 \cdot \frac{4}{9} = 4 \]
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Calculate \( 3\left(\frac{2}{3}\right) \): \[ 3 \cdot \frac{2}{3} = 2 \]
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Combine these results: \[ 4 + 2 - 5 = 6 - 5 = 1 \]
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The left side evaluates to 1, while the right side of the original equation is 6.
Since \( 1 \neq 6 \), we conclude that \( x = \frac{2}{3} \) is not a solution.
Thus, the correct explanation is:
A) No, \( x = \frac{2}{3} \) is not a solution because substituting it back into the equation results in \( 1 = 6 \), which is a false statement.
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