Solve question 1 in English completely showing all the steps used

Subtract $\frac{1}{x+2}$ from both sides:

The common denominator is $(1-2x)(x+2)$.

Combine the numerators:

These points indicate where the expression might change sign.

• Numerator: $4x + 3 = 0 \implies 4x = -3 \implies x = -\frac{3}{4}$
• Denominator: $1 - 2x = 0 \implies 1 = 2x \implies x = \frac{1}{2}$ $x + 2 = 0 \implies x = -2$

The critical points, in ascending order, are $x = -2$, $x = -\frac{3}{4}$, and $x = \frac{1}{2}$. These points divide the number line into four intervals: $(-\infty, -2)$, $(-2, -\frac{3}{4})$, $(-\frac{3}{4}, \frac{1}{2})$, and $(\frac{1}{2}, \infty)$.

• Interval 1: $(-\infty, -2)$ Test $x = -3$: $\frac{4(-3)+3}{(1-2(-3))(-3+2)} = \frac{-9}{(7)(-1)} = \frac{-9}{-7} = \frac{9}{7}$. Since $\frac{9}{7} > 0$, this interval is part of the solution.

• Interval 2: $(-2, -\frac{3}{4})$ Test $x = -1$: $\frac{4(-1)+3}{(1-2(-1))(-1+2)} = \frac{-1}{(3)(1)} = -\frac{1}{3}$. Since $-\frac{1}{3} \ngtr 0$, this interval is not part of the solution.

• Interval 3: $(-\frac{3}{4}, \frac{1}{2})$ Test $x = 0$: $\frac{4(0)+3}{(1-2(0))(0+2)} = \frac{3}{(1)(2)} = \frac{3}{2}$. Since $\frac{3}{2} > 0$, this interval is part of the solution.

• Interval 4: $(\frac{1}{2}, \infty)$ Test $x = 1$: $\frac{4(1)+3}{(1-2(1))(1+2)} = \frac{7}{(-1)(3)} = -\frac{7}{3}$. Since $-\frac{7}{3} \ngtr 0$, this interval is not part of the solution.

The solution is the union of the intervals $(-\infty, -2)$ and $(-\frac{3}{4}, \frac{1}{2})$.

The solution set is $\boxed{\left(-\infty, -2\right) \cup \left(-\frac{3}{4}, \frac{1}{2}\right)}$.