Browse Questions

# Solve the inequality for real $x 3(x-1) \leq 2(x-3)$

$\begin{array}{1 1} (A)\;(-6,\infty)\\(B)\;(-\infty,-3]\\(C)\;(-4,\infty]\\(D)\; (-9,\infty)\end{array}$

Toolbox:
• Same Quantity can be added (a subtracted ) to (from ) both sides of the inequality with out changing the sign of the in equality.
• Same positive quantities can be multiplied or divided to both side of the in equality with out changing the sign of the inequality.
• If same negative quantity is multiplied or divided to both sides of the inequality is reversed i.e $'>'$ sign changes to $'<'$ and $'<'$ changes $'>'$ .
The given inequality $3(x-1) \leq 2(x-3)$
=> $3(x-1) \leq 2x-6$
adding 3 on both sides of inequality
=> $3x-3+3 \leq 2x-6+3$
=> $3x \leq 2x-3$
Subtracting 2x on both sides,
$3x-2x \leq 2x-6+3$
=> $3x \leq 2x \leq 2x -3 -2x$
$=> x \leq -3$
Step 2:
All real numbers which are less then and equal to -3 satisfy the given in equality.
The solution set is $(- \infty, -3]$
Hence B is the correct answer.
edited Jul 25, 2014 by meena.p