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

Question

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

Answer 1

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.