# How do you solve 2x-3<x+4<3x-2?

Jul 27, 2015

$3 < x < 7$ (In interval notation: $\left(3 , 7\right)$)

#### Explanation:

In order for $2 x - 3 < x + 4 < 3 x - 2$ to be true, the compound inequality:

$2 x - 3 < x + 4$ $\text{ }$ AND $\text{ }$ $x + 4 < 3 x - 2$ must be true.

Solving each individually, we get:

$2 x - 3 < x + 4$ Subtract $x$ and add $3$ on both sides to get:

$x < 7$

We also need:

$x + 4 < 3 x - 2$ Subtract $x$ and add $2$ on both sides to get:

$6 < 2 x$, so $3 < x$

The solution will require both $3 < x$ and $x < 7$.

The numbers that satisfy both inequalities simultaneously are the numbers between $3$ and $7$ (exclusive).