Question

Find the solution set for the inequality $3x+9\ge -x+19$,

when the replacement set is a set of real numbers (R).

Answer 1

Given: $3x+9\ge –x+19$

$3x+9–9\ge –x+19–9$

$3x\ge –x+10$

$3x+x\ge –x+10+x$

$4x\ge 10$

Divide both sides by $4$, we get

$\frac{4x}{4}\ge \frac{10}{4}$

$x\ge \frac{5}{2}$

$\therefore$ The solution of the given inequality is,
$\frac{5}{2}\le x\le \infty$