The number of integral values of x satisfying the linear inequality |x|+|x+1| < 5 is .

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Solution

Given that |x|+|x+1| < 5.
Now, $(x | = (\begin{matrix} x, if x \geq 0 - x if x < 0 \end{matrix} (x + 1 | = (\begin{matrix} x + 1, if x \geq - 1 - x - 1 if x < - 1 \end{matrix} (x | + (x + 1 | = ⎛ ⎜ ⎝ \begin{matrix} x + x + 1 = 2 x + 1, if x \geq 0 - x + x + 1 = 1, if - 1 < x < 0 - x - x - 1 = - 2 x - 1, if x \leq - 1 \end{matrix} Now, (x | + (x + 1 | < 5 Case I : If x \geq 0 (x | + (x + 1 | = 2 x + 1 2 x + 1 < 5 \Rightarrow x < 2 \Rightarrow x = 0, 1 Case II : If - 1 < x < 0 (x | + (x + 1 | = 1 which is less than 5 . No integral value of x in this interval . Case III : If x \leq - 1 (x | + (x + 1 | = - 2 x - 1 - 2 x - 1 < 5 \Rightarrow - 2 x < 6 \Rightarrow x > - 3 ∴ x = - 2, - 1$
Therefore, the integral values of x satisfying the given linear inequation is -2, -1, 0, 1.
Number of values = 4

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