Number of integral values of $x$ satisfying the given inequality $1 \leq | 5 x + 8 | \leq 5$ is/are

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Solution

Given that $1 \leq | 5 x + 8 | \leq 5$

Using $a \leq | x | \leq b$ where $a, b > 0$
$\Rightarrow - b \leq x \leq - a$ or $a \leq x \leq b$

Here we have,
$1 \leq | 5 x + 8 | \leq 5 \Rightarrow - 5 \leq 5 x + 8 \leq - 1 or 1 \leq 5 x + 8 \leq 5$

$\Rightarrow - 13 \leq 5 x \leq - 9$ or $- 7 \leq 5 x \leq - 3$

$\Rightarrow - \frac{13}{5} \leq x \leq - \frac{9}{5}$ or $- \frac{7}{5} \leq x \leq - \frac{3}{5}$

$∴ x \in [- \frac{13}{5}, - \frac{9}{5}] \cup [- \frac{7}{5}, - \frac{3}{5}]$

The integral values of $x$ in above interval are ${- 2, - 1}$ .
Thus there are two integral values of $x$ satisfies the given inequality.

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