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Algebra · real student question

Solve the equation |3x + 4| = 2.

Question

Solve

∣3x+4∣=2\left|3x + 4\right| = 2

Step-by-step solution

  1. Check that a solution can exist at all. An absolute value is never negative, so ∣A∣=k|A| = k has solutions only when k≥0k \ge 0. Here k=2>0k = 2 > 0, so there will be exactly two — one on each side of the vertex.

  2. Split on the definition. ∣A∣=2|A| = 2 means AA sits 22 units from zero, in either direction:

    3x+4=2or3x+4=−23x + 4 = 2 \qquad \text{or} \qquad 3x + 4 = -2

    Writing both cases before touching either is what keeps the second solution from being lost.

  3. Solve the positive case.

    3x+4=2  ⟹  3x=−2  ⟹  x=−233x + 4 = 2 \;\Longrightarrow\; 3x = -2 \;\Longrightarrow\; x = -\frac{2}{3}

  4. Solve the negative case.

    3x+4=−2  ⟹  3x=−6  ⟹  x=−23x + 4 = -2 \;\Longrightarrow\; 3x = -6 \;\Longrightarrow\; x = -2

  5. Verify both and note the symmetry. ∣3(−23)+4∣=∣−2+4∣=2\left|3\left(-\tfrac23\right) + 4\right| = |-2 + 4| = 2 and ∣3(−2)+4∣=∣−6+4∣=2|3(-2) + 4| = |-6 + 4| = 2. Both roots sit symmetrically about the vertex x=−43x = -\tfrac43, at distance 23\tfrac23 on either side — the graph of ∣3x+4∣|3x+4| is a V with slope ±3\pm 3, and 23\tfrac{2}{3} is exactly the horizontal run needed for a rise of 22.

Answer

x=−23orx=−2x = -\frac{2}{3} \quad \text{or} \quad x = -2

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