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

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Answer 1 · 2015-07-27T21:20:10

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

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

$2 x - 3 < x + 4$ $2x-3 < x+4$ $" "$ AND $" "$ $x + 4 < 3 x - 2$ $x+4 < 3x-2$ must be true.

Solving each individually, we get:

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

$x < 7$ $x < 7$

We also need:

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

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

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

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

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