How many integer values of x satisfy both |2x - 3| < 9 and |x + 1| \ge 3? — GMAT

How many integer values of $x$ satisfy both $|2x - 3| < 9$ and $|x + 1| \ge 3$ ?

Answer: 4

There are 4 integer values of

x

that satisfy both inequalities.

Solving

|2x - 3| < 9

gives

-3 < x < 6

, so the set of possible integer values is

\{-2, -1, 0, 1, 2, 3, 4, 5\}

. Solving

|x + 1| \ge 3

gives

x \ge 2

x \le -4

. Taking the intersection of these two conditions gives

x \in \{2, 3, 4, 5\}

, for a total of 4 integer values.

Unfold the first absolute value inequality

|2x - 3| < 9

-9 < 2x - 3 < 9, which simplifies to -3 < x < 6

An absolute value inequality of the form |A| < B is equivalent to -B < A < B.

Unfold the second absolute value inequality

|x + 1| \ge 3

x + 1 \ge 3 or x + 1 \le -3, which simplifies to x \ge 2 or x \le -4

An absolute value inequality of the form |A| >= B is equivalent to A >= B or A <= -B.

List integer candidates and find the intersection

Candidates from first condition: {-2, -1, 0, 1, 2, 3, 4, 5}. Applying second condition (x >= 2 or x <= -4) leaves {2, 3, 4, 5}

The solution must satisfy both conditions simultaneously.

Solving systems of absolute value inequalities for integer solutions

Estimated Time:1m 30s

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