Solve the following systems of linear in equations:
$5 x-7<(x+3), 1-\frac{3 x}{2} \geq x-4$
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 $x \leq 2$