To solve the inequality $2|2x - 3| < |x + 10|$, we need to break it into different cases based on the expressions inside the absolute value signs.

Step 1: Set up the cases

The absolute value function creates different scenarios depending on whether the expression inside is positive or negative. For the expression $|2x - 3|$ and $|x + 10|$, we will examine the following cases:

Case 1: $2x - 3 \geq 0$ and $x + 10 \geq 0$

• Here, $|2x - 3| = 2x - 3$ and $|x + 10| = x + 10$.

The inequality becomes: $2(2x - 3) < x + 10$ Simplifying: $4x - 6 < x + 10$ Subtract $x$ from both sides: $3x - 6 < 10$ Add 6 to both sides: $3x < 16$ Divide by 3: $x < \frac{16}{3}$ Since this case assumes $2x - 3 \geq 0$, this implies $x \geq \frac{3}{2}$.

Therefore, the solution for this case is $\frac{3}{2} \leq x < \frac{16}{3}$.

Case 2: $2x - 3 \geq 0$ and $x + 10 < 0$

• Here, $|2x - 3| = 2x - 3$ and $|x + 10| = -(x + 10) = -x - 10$.

The inequality becomes: $2(2x - 3) < -x - 10$ Simplifying: $4x - 6 < -x - 10$ Add $x$ to both sides: $5x - 6 < -10$ Add 6 to both sides: $5x < -4$ Divide by 5: $x < -\frac{4}{5}$ However, this case assumes $x + 10 < 0$, meaning $x < -10$. The solution $x < -\frac{4}{5}$ does not satisfy $x < -10$, so there is no solution in this case.

Case 3: $2x - 3 < 0$ and $x + 10 \geq 0$

• Here, $|2x - 3| = -(2x - 3) = -2x + 3$ and $|x + 10| = x + 10$.

The inequality becomes: $2(-2x + 3) < x + 10$ Simplifying: $-4x + 6 < x + 10$ Add $4x$ to both sides: $6 < 5x + 10$ Subtract 10 from both sides: $-4 < 5x$ Divide by 5: $x > -\frac{4}{5}$ This case assumes $2x - 3 < 0$, which implies $x < \frac{3}{2}$.

Therefore, the solution for this case is $-\frac{4}{5} < x < \frac{3}{2}$.

Case 4: $2x - 3 < 0$ and $x + 10 < 0$

• Here, $|2x - 3| = -(2x - 3) = -2x + 3$ and $|x + 10| = -(x + 10) = -x - 10$.

The inequality becomes: $2(-2x + 3) < -x - 10$ Simplifying: $-4x + 6 < -x - 10$ Add $4x$ to both sides: $6 < 3x - 10$ Add 10 to both sides: $16 < 3x$ Divide by 3: [ x > \frac{16}{