**Question:**

Solve the following systems of linear in equations:

$5 x-7<(x+3), 1-\frac{3 x}{2} \geq x-4$

**Solution:**

When,

$5 x-7

Adding 7 to both the sides in the above equation

$5 x-7+7

$5 x

Now, subtracting x from both the sides

$5 x-x

$4 x<10$

Dividing both the sides by 4 in above equation

$\frac{4 x}{4}<\frac{10}{4}$

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

Now when,

$1-\frac{3 x}{2} \geq x-4$

Subtracting 1 from both the sides in the above equation

$1-\frac{3 x}{2}-1 \geq x-4-1$

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

Now multiplying both the sides by 2 in the above equation

2. $\left(\frac{-3 x}{2}\right) \geq 2 x-10$

$-3 x \geq 2 x-10$

Now subtracting 2x from both the sides in the above equation

$-3 x-2 x \geq 2 x-10-2 x$

$-5 x \geq-10$

Now, multiplying both the sides by -1 in the above equation

$-5 x(-1) \geq-10(-1)$

$5 x \leq 10$

Now, dividing both the sides by 5 in the above equation

$\frac{5 x}{5} \leq \frac{10}{5}$

$x \leq 2$

Therefore,

$x<\frac{5}{2}$ and $x \leq 2$