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Exercise 5.1 · Q12

Q.12(3x5+4)≥13(x−6)\dfrac{1}{2}\left(\dfrac{3x}{5} + 4\right) \ge \dfrac{1}{3}(x - 6)

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Clearing the fractions by multiplying through by the LCM of the denominators, then isolating xx, gives the solution x≤120x \le 120.

Why This Approach Works

When an inequality has several fractional terms, the cleanest first move is to clear every denominator at once by multiplying through by their least common multiple (LCM). Since 22, 55 and 33 have LCM 3030, and 3030 is positive, multiplying by it never flips the inequality direction.

Watch out

Every term on BOTH sides must be multiplied by the LCM -- not just the fractions. Skipping a constant term is a common source of error here.


Step-by-Step Solution

1. Write the inequality clearly.

12(3x5+4)≥13(x−6)\frac{1}{2}\left(\frac{3x}{5} + 4\right) \ge \frac{1}{3}(x - 6)

2. Multiply both sides by 30 (the LCM of 2, 5, 3) to eliminate every fraction.

30⋅12(3x5+4)≥30⋅13(x−6)30 \cdot \frac{1}{2}\left(\frac{3x}{5} + 4\right) \ge 30 \cdot \frac{1}{3}(x - 6)

15(3x5+4)≥10(x−6)15\left(\frac{3x}{5} + 4\right) \ge 10(x - 6)

3. Distribute on both sides.

Left: 15⋅3x5=9x15 \cdot \frac{3x}{5} = 9x, and 15⋅4=6015 \cdot 4 = 60, so the left side is 9x+609x + 60.

Right: 10(x−6)=10x−6010(x-6) = 10x - 60.

9x+60≥10x−609x + 60 \ge 10x - 60

4. Bring variable terms to one side, constants to the other. …

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