Solve the following systems of linear in equations:

Solution:

When

$\frac{7 x-1}{2}<-3$

Multiplying both the sides by 2

$\left(\frac{7 x-1}{2}\right)(2)<-3(2)$

$7 x-1<-6$

Adding 6 to both the sides in above equation

$7 x-1+6<-6+6$

$7 x+5<0$

Subtracting 5 from both the sides in above equation

$7 x+5-5<0-5$

$7 x<-5$

Dividing both the sides by 7 in above equation

$\frac{7 x}{7}<\frac{-5}{7}$

Therefore

$x<\frac{-5}{7}$

Now when,

$\frac{3 x+8}{5}+11<0$

Subtracting both the sides by 11 in the above equation

$\frac{3 x+8}{5}+11-11<0-11$

$\frac{3 x+8}{5}<-11$

Multiplying both the sides by 5 in the above equation

$\left(\frac{3 x+8}{5}\right)(5)<-11(5)$

$3 x+8<-55$

Subtracting 8 from both the sides in above equation

$3 x+8-8<-55-8$

$3 x<-63$

Dividing both the sides by 3 in above equation

$\frac{3 x}{3}<\frac{-63}{3}$

Therefore,

$x<-21$