(10-х) *(3х+4) >0

To solve the inequality \((10 - x)(3x + 4) > 0\), we need to determine where the product of the two factors is positive. We can start by identifying the zeros of each factor. 1. **Find the zeros of each factor**: - For \(10 - x = 0\): \[ x = 10 \] - For \(3x + 4 = 0\): \[ 3x = -4 \implies x = -\frac{4}{3} \] 2. **Identify critical points**: The critical points from our factors are \(x = 10\) and \(x = -\frac{4}{3}\). These points divide the number line into three intervals: - \( (-\infty, -\frac{4}{3}) \) - \( (-\frac{4}{3}, 10) \) - \( (10, \infty) \) 3. **Test each interval**: We will select test points from each interval and check the sign of the product \((10 - x)(3x + 4)\). - **Interval \( (-\infty, -\frac{4}{3}) \)**: Choose \(x = -2\) \[ (10 - (-2))(3(-2) + 4) = (10 + 2)(-6 + 4) = 12 \times (-2) = -24 \quad (\text{negative}) \] - **Interval \( (-\frac{4}{3}, 10) \)**: Choose \(x = 0\) \[ (10 - 0)(3(0) + 4) = 10 \times 4 = 40 \quad (\text{positive}) \] - **Interval \( (10, \infty) \)**: Choose \(x = 11\) \[ (10 - 11)(3(11) + 4) = (-1)(33 + 4) = -1 \times 37 = -37 \quad (\text{negative}) \] 4. **Conclusion**: The product \((10 - x)(3x + 4)\) is positive in the interval: \[ (-\frac{4}{3}, 10) \] Since we want the inequality \((10 - x)(3x + 4) > 0\), we state the solution as: \[ x \in \left(-\frac{4}{3}, 10\right) \] Thus, the final answer is: \[ (-\frac{4}{3}, 10) \]