To determine which equation has the solution set \( x = {-95, 3} \), we can set each equation equal to zero and solve for \( x \).
-
Equation: \((x - 3)(5x + 9) = 0\)
- Solutions:
- \(x - 3 = 0 \rightarrow x = 3\)
- \(5x + 9 = 0 \rightarrow 5x = -9 \rightarrow x = -\frac{9}{5}\)
- Solution set: {3, -\frac{9}{5}}
- Solutions:
-
Equation: \(-3x(5x + 9) = 0\)
- Solutions:
- \(-3x = 0 \rightarrow x = 0\)
- \(5x + 9 = 0 \rightarrow 5x = -9 \rightarrow x = -\frac{9}{5}\)
- Solution set: {0, -\frac{9}{5}}
- Solutions:
-
Equation: \((x + 3)(9x - 5) = 0\)
- Solutions:
- \(x + 3 = 0 \rightarrow x = -3\)
- \(9x - 5 = 0 \rightarrow 9x = 5 \rightarrow x = \frac{5}{9}\)
- Solution set: {-3, \frac{5}{9}}
- Solutions:
-
Equation: \((x - 3)(9x + 5) = 0\)
- Solutions:
- \(x - 3 = 0 \rightarrow x = 3\)
- \(9x + 5 = 0 \rightarrow 9x = -5 \rightarrow x = -\frac{5}{9}\)
- Solution set: {3, -\frac{5}{9}}
- Solutions:
None of the provided equations gives a solution set of \({-95, 3}\).
If \( x = -95 \) is to be included in the solution set, let's analyze what that would require:
- Something like \((x + 95)(x - 3) = 0\) would give \(x = -95\) and \(x = 3\). But none of the available options produce this equation.
Thus, it appears there is no solution set among the options that equals \({-95, 3}\).