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Algebra · real student question

Solve the inequality (2x - 5)/4 < (3x - 3)/3.

Question

Solve

2x−54<3x−33\frac{2x-5}{4} < \frac{3x-3}{3}

Step-by-step solution

  1. Multiply by the least common multiple of the denominators. lcm⁡(4,3)=12\operatorname{lcm}(4,3) = 12. Because 12>012 > 0, the direction of the inequality is preserved — this is the one place an inequality differs from an equation, and it only becomes an issue when the multiplier is negative.

    12⋅2x−54<12⋅3x−33⟹3(2x−5)<4(3x−3)12 \cdot \frac{2x-5}{4} < 12 \cdot \frac{3x-3}{3} \quad\Longrightarrow\quad 3(2x-5) < 4(3x-3)

  2. Expand both sides.

    6x−15<12x−126x - 15 < 12x - 12

  3. Collect the x terms on the side with the larger coefficient. Subtracting 6x6x keeps the coefficient positive, so no sign flip is needed later:

    −15<6x−12-15 < 6x - 12

  4. Isolate x.

    −15+12<6x⟹−3<6x⟹x>−12-15 + 12 < 6x \quad\Longrightarrow\quad -3 < 6x \quad\Longrightarrow\quad x > -\frac{1}{2}

  5. Test the boundary and one point on each side. At x=−12x = -\tfrac12: left =−64=−1.5= \tfrac{-6}{4} = -1.5, right =−4.53=−1.5= \tfrac{-4.5}{3} = -1.5 — equal, so the endpoint is correctly excluded. At x=0x = 0: −1.25<−1-1.25 < -1, true. At x=−1x = -1: −1.75<−2-1.75 < -2, false. The solution set is (−12, ∞)\left(-\tfrac12,\ \infty\right).

Answer

x>−12,i.e. x∈(−12, ∞)x > -\frac{1}{2}, \qquad \text{i.e. } x \in \left(-\tfrac{1}{2},\ \infty\right)

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