Solve the following systems of linear in equations:

Solution:

$3 x-x>x+\frac{4-x}{3}$ and $x+\frac{4-x}{3}>3$

When,

$3 x-x>x+\frac{4-x}{3}$

$2 x>x+\frac{4-x}{3}$

Subtracting x from both the sides in above equation

$2 x-x>x+\frac{4-x}{3}-x$

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

Multiplying both the sides by 3 in the above equation

$3 x>3^{\left(\frac{4-x}{3}\right)}$

$3 x>4-x$

Adding x on both the sides in above equation

$3 x+x>4-x+x$

$4 x>4$

Dividing both the sides by 4 in above equation

$\frac{4 x}{4}>\frac{4}{4}$

$x>1$

Now when

$x+\frac{4-x}{3}>3$

Multiplying both the sides by 3 in above equation

$3 x+3\left(\frac{4-x}{3}\right)>3(3)$

$3 x+4-x>9$

$2 x+4>9$

Subtracting 4 from both the sides in above equation

$2 x+4-4>9-4$

$2 x>5$

Dividing both the sides by 2 in above equation

$\frac{2 x}{2}>\frac{5}{2}$

$x>\frac{5}{2}$

Merging overlapping intervals

$x>\frac{5}{2}$

Therefore

$x \in\left(\frac{5}{2}, \infty\right)$