waecmaths question:
Solve the inequality $3(x+1)\le 5(x+2)+15$
Option A:
$x\ge -14$
Option B:
$x\le -14$
Option C:
$x\le -11$
Option D:
$x\ge -11$
waecmaths solution:
$\begin{align} & 3(x+1)\le 5(x+2)+15 \\ & 3x+3\le 5x+10+15 \\ & 3x+3\le 5x+25 \\ & -2x\le 22 \\ & x\ge -11 \\\end{align}$
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