Question

What is the solution set for $\frac{2(x-1)}{5}\le \frac{3(2+x)}{7}?$

• A. $(-24,\infty )$
• B. $(-44,\infty )$
• C. $(-4,\infty )$
• D. $(-12,\infty )$

Answer 1

The correct option is B $(-44,\infty )$

$\frac{2(x-1)}{5}\le \frac{3(2+x)}{7}$
$14(x-1)\le 15(2+x)$
$14x-14\le 30+15x$
$-x\le 44$
$x\ge -44$
or $x$ lies in $(-44,\infty )$