Jambmaths question:
Solve for x in $\frac{4x-6}{3}\le \frac{3+2x}{2}$
Option A:
$x\le 10\tfrac{1}{2}$
Option B:
$x\le 20\tfrac{1}{2}$
Option C:
$x\ge 20\tfrac{1}{2}$
Option D:
$x>10\tfrac{1}{2}$
Jamb Maths Solution:
$\begin{align} & \frac{4x-6}{3}\le \frac{3+2x}{2} \\ & \text{Multiply both sides by 6} \\ & \frac{6(4x-6)}{3}\le \frac{6(3+2x)}{2} \\ & 2(4x-6)\le 3(3+2x) \\ & 8x-12\le 9+6x \\ & 8x-6x\le 9+12 \\ & 2x\le 21 \\ & x\le \frac{21}{2} \\\end{align}$
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