Question

Solve the inequality for $x:$
$\frac{x+1}{3}<\frac{x-2}{4}<x$

Answer 1

Given inequality: $\frac{x+1}{3}<\frac{x-2}{4}<x$

The given inequality can be broken as:
$\frac{x+1}{3}<\frac{x-2}{4}\&\frac{x-2}{4}<x$
$\Rightarrow \frac{x+1}{3}<\frac{x-2}{4}\cdots \left(i\right)$
$\&\frac{x-2}{4}<x\cdots \left(ii\right)$
Now, Solving the first inequality, we get:
$\frac{x+1}{3}<\frac{x-2}{4}$

$\Rightarrow 4x+4<3x-6$

$\Rightarrow 4x-3x<-6-4$

$\Rightarrow x<-10\cdots \left(a\right)$

And, similarly solving the second inequality
$\frac{x-2}{4}<x$

$\Rightarrow x-2<4x\Rightarrow x-4x<2$

$\Rightarrow -3x<2\Rightarrow x>-\frac{2}{3}\cdots \left(b\right)$

And for the given inequality to be satisfied, the equation $\left(a\right)$ and equation $\left(b\right)$ must be simultaneously satisfied.

Thus, from $\left(a\right)\&\left(b\right),$ we get:


We can see from the above diagram that No common solution exists.
Hence the option (d) is correct answer.