Question

Solve the following system of equations in R.
$|3-4x|\ge 9$

Answer 1

$|3-4x|\ge 9\Rightarrow 4|\frac{3}{4}-x|\ge 9\Rightarrow |\frac{3}{4}-x|\ge \frac{9}{4}CaseI:When-\infty <x\le -\frac{3}{4}|\frac{3}{4}-x|=(\frac{3}{4}-x)\therefore |\frac{3}{4}-x|\ge \frac{9}{4}\Rightarrow \therefore (\frac{3}{4}-x)\ge \frac{9}{4}\Rightarrow -\frac{6}{4}\ge x\Rightarrow -\frac{3}{2}\ge xBut,-\infty <x<-1.\therefore \text{The solution set of the given inequation is}[-\infty ,-\frac{3}{2}].Case2:When-\frac{3}{4}<x<\infty |\frac{3}{4}-x|=-[\frac{3}{4}-x]\therefore |\frac{3}{4}-x|\ge \frac{9}{4}\Rightarrow -[\frac{3}{4}-x]\ge \frac{9}{4}\Rightarrow x\ge 3But,-\frac{3}{4}<x<\infty \therefore \text{The solution set of the given inequation is}[3,\infty ]\text{Combining case 1 and case 2,we obtain that the solution set of given in inequality is}[-\infty ,\frac{3}{2}]\cup [3,\infty ]xϵ(\frac{3}{4},1)\text{Combining (ii) and (iii) we get}(\frac{3}{4},1)\cup (1,\infty )\text{as the solution set.}$