To solve the factored quadratic equation \((19x-3)(9+x) = 0\), we need to find the values of \(x\) that make either of the factors equal to zero.
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Set the first factor to zero: \[ 19x - 3 = 0 \] Solving for \(x\), we add \(3\) to both sides: \[ 19x = 3 \] Now divide by \(19\): \[ x = \frac{3}{19} \]
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Set the second factor to zero: \[ 9 + x = 0 \] Solving for \(x\), we subtract \(9\) from both sides: \[ x = -9 \]
Thus, the solution set for the equation \((19x-3)(9+x) = 0\) is: \[ x = {-9, \frac{3}{19}} \]
Now, we can check the provided options against the solutions we derived.
- The solution set is \(x={ -9,\frac {19}{3}}\)
- The solution set is \(x={ -9,\frac {3}{10}}\)
- The solution set is \(x={ \frac {19}{3},9}\)
- The solution set is \(x={ -9,3}\)
Among these, \( \frac{19}{3} \) and \( \frac{3}{10} \) are not solutions based on our calculations, nor is \(9\) or \(3\). Therefore, the provided solution set is incorrect.
Final conclusion: The correct solution set is \(x = {-9, \frac{3}{19}}\).