Question

If $|2x-3|<|x+5|,$ then $x$ lies in the interval

• A. $(-3,5)$
• B. $(5,9)$
• C. $\left(-\frac{2}{3},8\right)$
• D. $\left(-8,\frac{2}{3}\right)$
• E. $\left(-5,\frac{2}{3}\right)$

Answer 1

Answer: (C)

$|2x-3|<|x+5|$
$\Rightarrow$ $\pm (2x-3)<\pm (x+5)$ ?.. (i) First we take positive sign and negative sign on both sides,
$\Rightarrow$ $2x-3<x+5$
$\Rightarrow$ $x<8$ ?.(ii) $-(2x-3)<-(x+5)$
$\Rightarrow$ $2x-3>x+5$
$\Rightarrow$ $x>8$ ..(iii)
Now, we take negative sign in RHS and positive in LHS,
$\Rightarrow$ $2x-3<-x-5$
$\Rightarrow$ $3x<-2$
$\Rightarrow$ $x<-\frac{2}{3}$ .. (iv)
and we take negative sign LHS and positive in RHS,
$\Rightarrow$ $-(2x-3)<x+5$
$\Rightarrow$ $-2x+3<x+5$
$\Rightarrow$ $-2<3x$
$\Rightarrow$ $-2/3<x$ ?.(v)
From Eqs. (ii) and (v), $-2/3<x<8$
$\Rightarrow$ $x\in (-2/3,8)$
From Eqs. (iii) and (iv), $x\in (-\infty ,-2/3)\cup (8,\infty )$