Solve |2x + 0| ≥ 3
Absolute value inequality, worked out line by line the way a teacher would write it.
Answer
| Solution | x ≤ −3/2 or x ≥ 3/2 |
| Interval notation | (−∞, −3/2] ∪ [3/2, ∞) |
Step-by-step solution
11 steps-
1 Given\left|{2 x + 0}\right| \ge 3
Solve for x.
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2 Isolate the absolute value\left|{x}\right| \ge \frac{3}{2}
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3 |u| > d means u is below −d or above d
|u| is the distance from u to 0. Being more than 3/2 away from 0 means u is below −3/2 or above 3/2, so the answer has two separate pieces.
x \le - \frac{3}{2}\quad\text{or}\quad x \ge \frac{3}{2} -
4 Case 1x \le - \frac{3}{2}
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5 Multiply both sides by 2 to clear the fractions
2 is the least common multiple of the denominators. Multiplying every term on both sides by it cancels each fraction, so the rest of the work uses whole numbers. The equation stays balanced because both sides were multiplied by the same number.
2 x \le -32 is positive, so the inequality sign stays the same.
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6 Divide both sides by 2x \le - \frac{3}{2}
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7 Case 2x \ge \frac{3}{2}
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8 Multiply both sides by 2 to clear the fractions2 x \ge 3
2 is positive, so the inequality sign stays the same.
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9 Divide both sides by 2x \ge \frac{3}{2}
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10 Either case works, so combine the two answersx \le - \frac{3}{2} \;\text{ or }\; x \ge \frac{3}{2}
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11 Solutionx \le - \frac{3}{2} \;\text{ or }\; x \ge \frac{3}{2}\qquad \left( -\infty, - \frac{3}{2} \right] \cup \left[ \frac{3}{2}, \infty \right)
Check with test values
| x | Left side | Right side | Holds? |
|---|---|---|---|
| −2 | 4 | 3 | ✓ true |
| 2 | 4 | 3 | ✓ true |
| −1 | 2 | 3 | ✗ false |
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