$\begin{array}{1 1}(A)\;[\frac{-80}{3},\frac{-10}{3}]\\(B)\;[-4,2]\\(C)\;(-23,2]\\(D)\;[2,3]\end{array} $

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- 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 $'>'$ .

Given inequality is $ -12 \leq 4 - \large\frac{3x}{-5}$$ 2$

Adding -4 throughout the inequality

$=> 12 - 4 < 4 - \large\frac{-3x}{-5} $$-4 \leq 2-4$

$=> -16 < \large\frac{-3x}{-5} $$ \leq -2$

$=> -16 < \large\frac{3x}{5} $$ \leq -2$

Multiplying by 5 through out

$=> -80 <3x \leq -10$

Dividing by 3 through out

=> $ \large\frac{-80 }{3} $$ < x \leq \large\frac{-10}{3}$

Step 2:

All numbers greater than $ \large\frac{-80}{3}$ and less than or equal to $ \large\frac{-10}{3}$ are solutions to the given inequality

The solution set is $[\large\frac{-80}{3}, \frac{-10}{3} ]$

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