Question

1. $|4x - 3|-xgeq15$

Explanation:

Step1: Consider two cases for the absolute - value.

Case 1: When $4x - 3\geq0$ (i.e., $x\geq\frac{3}{4}$), the inequality becomes $(4x - 3)-x\geq15$.
Simplify the left - hand side: $4x-3 - x=3x - 3$. So, $3x-3\geq15$.
Add 3 to both sides: $3x\geq15 + 3$, which gives $3x\geq18$.
Divide both sides by 3: $x\geq6$. Since $6\geq\frac{3}{4}$, this solution is valid for this case.
Case 2: When $4x - 3\lt0$ (i.e., $x\lt\frac{3}{4}$), the inequality becomes $-(4x - 3)-x\geq15$.
Expand the left - hand side: $-4x + 3-x\geq15$.
Combine like terms: $-5x+3\geq15$.
Subtract 3 from both sides: $-5x\geq15 - 3$, so $-5x\geq12$.
Divide both sides by $- 5$ and reverse the inequality sign: $x\leq-\frac{12}{5}$. Since $-\frac{12}{5}\lt\frac{3}{4}$, this solution is valid for this case.

Answer:

$x\leq-\frac{12}{5}$ or $x\geq6$