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Exercise 5.1 · Q20

Q.x2≥5x−23−7x−35\dfrac{x}{2} \ge \dfrac{5x - 2}{3} - \dfrac{7x - 3}{5}

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Clear denominators with the LCM 3030, simplify, and solve. The solution is x≥−27x \ge -\dfrac{2}{7}, i.e. [−27,∞)\left[-\dfrac{2}{7}, \infty\right).

This is a linear inequality in one variable. Multiplying every term by the LCM of the denominators removes the fractions; the multiplier is positive, so the inequality direction is preserved.

Step-by-step solution

  1. Find the LCM of 22, 33, 55. It is 3030.

  2. Multiply every term by 3030.

    • 30⋅x2=15x30\cdot\dfrac{x}{2} = 15x
    • 30⋅5x−23=10(5x−2)30\cdot\dfrac{5x-2}{3} = 10(5x-2)
    • 30⋅7x−35=6(7x−3)30\cdot\dfrac{7x-3}{5} = 6(7x-3)

    The inequality becomes:

15x≥10(5x−2)−6(7x−3)15x \ge 10(5x-2) - 6(7x-3)

  1. Expand. Note −6×(−3)=+18-6\times(-3) = +18:

15x≥50x−20−42x+1815x \ge 50x - 20 - 42x + 18

  1. Combine like terms on the right. 15x≥8x−215x \ge 8x - 2 …

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