Question

Solve the following inequation:

$13x-5<15x+4<7x+12,x\in R$.

• A. $\{x:x\in R\text{and}-4.5<x\le 1\}$
• B. $\{x:x\in R\text{and}-4.5\le x<1\}$
• C. $\{x:x\in R\text{and}-4.5\le x\le 1\}$
• D. $\{x:x\in R\text{and}-4.5<x<1\}$

Answer 1

The correct option is D $\{x:x\in R\text{and}-4.5<x<1\}$

Given: $13x-5<15x+4<7x+12$

$\Rightarrow 13x-5<15x+4\text{and}15x+4<7x+12$

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 above inequations.

$\Rightarrow -5-4<15x-13x\text{and}15x-7x<12-4$

$\Rightarrow -9<2x\text{and}8x<8$

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.
On dividing left hand side inequation by sides by 2 and right hand side inequation by 8, we get:

$\Rightarrow -4.5<x\text{and}x<1$

Since $x\in R$,

Solution set is given by, $\{x:x\in Rand-4.5<x<1\}$