Question

Find the solution set for the inequality $3j-4\le 6j+11$.

• A. $j\le -5$
• B. $j\ge -5$
• C. $j\le 5$
• D. $j\ge 5$

Answer 1

The correct option is B $j\ge -5$

We need to find solution of $3j-4\le 6j+11$

Add $-6j$ and $4$ both sides, we get

$3j-4-6j+4\le 6j+11-6j+4$

$\Rightarrow -3j\le 15$

Multiply $-\frac{1}{3}$ on both sides, we get

$-3j\times (-\frac{1}{3})\ge 15\times -\frac{1}{3}$

$\Rightarrow j\ge -5$