Math · College algebra · Worked example
Solve an absolute value inequality
Solve |3x + 1| ≥ 5 and write the answer in interval notation.
Read it as a distance
The absolute value is already isolated. |3x + 1| ≥ 5 says 3x + 1 is at least 5 units from zero, so it is at most −5 or at least 5: two pieces joined by “or”.
Solve each piece
Subtract 1 and divide by 3, a positive number, so neither sign changes: 3x ≤ −6 gives x ≤ −2, and 3x ≥ 4 gives x ≥ 4/3.
Write the union
The solution is two rays. Both endpoints are included because the original sign is ≥, so they take brackets, and ∪ joins the pieces.
Check on the graph
The graph of y = |3x + 1| meets the line y = 5 at x = −2 and x = 4/3. The V is on or above the line left of −2 and right of 4/3, and below it in between. A test value agrees: x = 0 gives |1| = 1, less than 5, so 0 is rightly left out.
Result
x ≤ −2 or x ≥ 4/3, which is (−∞, −2] ∪ [4/3, ∞).
Your turn
Solve |2x − 1| < 5 and write the answer in interval notation.
Show the answer and explanation
−2 < x < 3, which is (−2, 3).
“Less than” gives one interval: −5 < 2x − 1 < 5. Add 1 to all three parts: −4 < 2x < 6. Divide by 2: −2 < x < 3. Both endpoints are left out because the sign is strict.
Keep exploring
In Graph, look where the V lies below the line instead: that gap, −2 < x < 4/3, is the solution of |3x + 1| < 5.
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