Question

How many different integers values of $x$ satisfy $\left| {x + 6} \right| < 3$?

Answer 1

To solve this question, we need to have a basic idea about absolute value function.
If x is an absolute function, then we have the relation $|x|= +x$ for $x>0$ and $|x|=-x$ for $x<0$.
In absolute value inequalities, for example if $|x|<1$ then we can write two cases
case 1: $+x < 1$
case 2: $-x < 1$ $\Rightarrow x >-1$
Using this information, we will simplify the given inequality and get the required solution.

Complete step by step answer:
We have given an inequality,
$\left| {x + 6} \right| < 3$
We have to find the number of different integers which will satisfy the given inequality.
In order to solve this we open mod by positive and negative both then we solve inequality and find intersection.
Part 1: $x + 6 < 3$ when absolute value is positive.
Part 2: $x + 6 > - 3$ when absolute value is negative.

Part 1: Solving the positive part,
$x + 6 < 3$ when absolute value is positive
On subtracting 6 from both sides,
$x+6-6 < 3 - 6$
On simplifying inequality.
$x < \- 3$ ……(i)

Part 2: When absolute value is negative.
$x + 6 > - 3$
On subtracting 6 on both sides
$x +6-6 > - 3 - 6$
On simplifying inequality.
$x > - 9$ ……(ii)

On taking intersection from inequality (i) and (ii)
$- 9 < x < \- 3$
Integers between this inequality are $- 4, - 5, - 6, - 7, - 8$
Therefore, We got 5 integer solutions satisfying the $\left| {x + 6} \right| < 3$.

Note:
To solve these types of questions students must know the concept of inequality. Students must know when the modulus(absolute) function is open with a positive sign and when the mod is open with a negative sign. When we multiply inequality with negative then the inequality is changed. And if we multiply with positive then that inequality remains the same. Students often make mistakes in opening the sign of the modulus function.