Question

Помогите пожалуйста Решите неравенство!!!

Answer 1

$\frac{x^2+2x-3}{2x-3} \geq 0\; ,\; \; 2x-3\ne 0\\\\ \frac{(x-1)(x+3)}{2(x-\frac{3}{2})} \geq 0\\\\---[-3\, ]+++[\, 1\, ]---(\frac{3}{2})+++\\\\x\in [-3,1\, ]\cup (\frac{3}{2},+\infty )$

Answer 2

Solving the Inequality

To solve the given inequality, we need to find the values of x that satisfy the inequality. The inequality provided is not clear, so I will assume that the inequality is 8 + 4x > 0 and 4 - 3x > -8.

Let's solve each inequality separately:

1. 8 + 4x > 0: - Subtract 8 from both sides: 4x > -8. - Divide both sides by 4: x > -2.

2. 4 - 3x > -8: - Add 8 to both sides: 4 + 8 > 3x. - Simplify: 12 > 3x. - Divide both sides by 3: 4 > x.

Therefore, the solution to the system of inequalities is x > -2 and x < 4.

Please let me know if you have any further questions!