Solve the following systems of linear in equations:

Solution:

$-12<4-\frac{3 x}{-5}$ and $4-\frac{3 x}{-5} \leq 2$

When,

$-12<4-\frac{3 x}{-5}$

$4-\frac{3 x}{-5}>-12$

Subtracting 4 from both the sides in above equation

$4-\frac{3 x}{-5}-4>12-4$

$-\frac{\frac{3 x}{-5}} >-16$

$\frac{3 x}{5}>-16$

Multiplying both the sides by 5 in the above equation

$\left(\frac{3 x}{5}\right)(5)>-16(5)$

$3 x>-80$

Dividing both the sides by 3 in above equation

$\left(\frac{3 x}{3}\right)>\frac{-80}{3}$

Therefore,

$x>\frac{-80}{3}$

Now when,

$4-\frac{3 x}{-5} \leq 2$

Subtracting both the sides by 4 in the above equation

$4-\frac{\frac{3 x}{-5}}{-4} \leq 2-4$

$-\frac{3 x}{-5} \leq-2$

$\frac{3 x}{5} \leq-2$

Multiplying both the sides by 5 in the above equation

$3 x \leq-10$

Dividing both the sides by 3 in the above equation

$\frac{3 x}{3} \leq \frac{-10}{3}$

Therefore,

$x \leq \frac{-10}{3}$

Therefore: $\mathrm{x} \in\left(\frac{-80}{3}, \frac{-10}{3}\right]$