Question

Solve the following inequation:

$4x-19<\frac{3x}{5}-2\le \frac{-2}{5}+x,x\in R$

• A. $-4\le x<5$
• B. $4\le x<5$
• C. $-4\le x\le 5$
• D. $-4<x<5$

Answer 1

The correct option is A $-4\le x<5$

Given: $4x-19<\frac{3x}{5}-2\le \frac{-2}{5}+x$

The given inequation can be written as;
$\Rightarrow 4x-19<\frac{3x}{5}-2\text{and}\frac{3x}{5}-2\le \frac{-2}{5}+x$

Rule: If a term of an inequation is transferred from one side to the other side of the inequation, the sign of the term gets changed. Let's apply this rule in the following inequations.

$\Rightarrow 4x<\frac{3x}{5}-2+19\text{and}\frac{3x}{5}-x\le \frac{-2}{5}+2$

$\Rightarrow 4x-\frac{3x}{5}<17\text{and}\frac{3x}{5}-x\le 2-\frac{2}{5}$

$\Rightarrow \frac{17x}{5}<17\text{and}\frac{-2x}{5}\le \frac{8}{5}$

Rule: If both the sides of an inequation are multiplied or divided by the same positive number, then the sign of the inequality will remain the same.

Rule: If both the sides of an inequation are multiplied or divided by the same negative number, then the sign of the inequality will get reversed.

let's apply the above rules in following ways. On multiplying the Left hand side inequation by 5 and right hand side inequation by-1, we get

$\Rightarrow x<5andx\ge -4$

$\therefore$ solution set is $-4\le x<5$.