Solve the following inequalities: (a) 2 - 5x 9 + 2x (b) |3 + 2t| 6 (c) x - 1 / 3x + 5 > 0 (d) (3t + 2)2 > 16 (e) 2x2 − x − 3 < 0

Remember to reverse the inequality sign when dividing by a negative number.

The solution is $\boxed{\text{}x \ge -1\text{}}$

Q2 b)

The solution is $\boxed{\text{} -\frac{9}{2} \le t \le \frac{3}{2}\text{}}$

Q2 c)

For the numerator: $x - 1 = 0 \implies x = 1$
For the denominator: $3x + 5 = 0 \implies 3x = -5 \implies x = -\frac{5}{3}$

• For $x < -\frac{5}{3}$ (e.g., $x = -2$): $\frac{-2 - 1}{3(-2) + 5} = \frac{-3}{-1} = 3 > 0$. This interval is a solution.
• For $-\frac{5}{3} < x < 1$ (e.g., $x = 0$): $\frac{0 - 1}{3(0) + 5} = \frac{-1}{5} < 0$. This interval is not a solution.
• For $x > 1$ (e.g., $x = 2$): $\frac{2 - 1}{3(2) + 5} = \frac{1}{11} > 0$. This interval is a solution. The solution is $\boxed{\text{}x < -\frac{5}{3} \text{ or } x > 1\text{}}$

Q2 d)

Remember to use the absolute value.

For the first inequality:

For the second inequality:

The solution is $\boxed{\text{}t < -2 \text{ or } t > \frac{2}{3}\text{}}$

Q2 e)

We can factor the quadratic:

Setting each factor to zero gives the roots:

Since the coefficient of $x^2$ (which is 2) is positive, the parabola opens upwards. This means the quadratic expression $2x^2 - x - 3$ is negative between its roots.
The roots are $x = -1$ and $x = \frac{3}{2}$.
The solution is $\boxed{\text{} -1 < x < \frac{3}{2}\text{}}$