Question

x/4 < (5x-2)/3 - (7x-3)/5​

Answer 1

Question :

Solve the in-equation

$\sf\dfrac{x}{4}<\:\dfrac{(5x-2)}{3}-\dfrac{(7x-3)}{5}$ x ε R

Theory :

Solution of linear equations :

1. Same number may be added to (or subtracted from ) both sides of an inequation without changing the sign of inequality.
2. The sign of inequality is reversed when both sides of an inequation are multipled or divided by a negative number .

Solution :

We have ,

$\sf\dfrac{x}{4}<\:\dfrac{(5x-2)}{3}-\dfrac{(7x-3)}{5}$

$\sf\dfrac{x}{4}<\:\dfrac{5(5x-2)-3(7x-3)}{15}$

$\sf\dfrac{x}{4}<\:\dfrac{(25x-10-21x+9)}{15}$

$\sf\dfrac{x}{4}<\:\dfrac{(4x-1)}{15}$

Now multiply both sides by 15×4 , then

$\sf\:(15\times4)\dfrac{x}{4}<\:\dfrac{4x-1}{15}(15\times4)$

$\sf\:15x<4(4x-1)$

$\sf\:15x<16x-4$

$\sf\:4<16x-15x$

$\sf\:4<x$

It is the required solution !

Answer 2

Solution:- Refer In Attachment..