Question 8 of 40intermediate🔧 ApplyNumerical2 marks

Solve the inequality for real xx: 12(3x5+4)≥13(x−6)\frac{1}{2}\left(\frac{3x}{5} + 4\right) \geq \frac{1}{3}(x - 6)

Correct Answer

We start with the inequality 12(3x5+4)≥13(x−6)\frac{1}{2}\left(\frac{3x}{5} + 4\right) \geq \frac{1}{3}(x - 6). Simplifying the expressions on both sides, we get 3x10+2≥x3−2\frac{3x}{10} + 2 \geq \frac{x}{3} - 2.

To clear the denominators, we multiply both sides by the LCM of 10 and 3, which is 30. This gives 30(3x10+2)≥30(x3−2)30\left(\frac{3x}{10} + 2\right) \geq 30\left(\frac{x}{3} - 2\right), which simplifies to 9x+60≥10x−609x + 60 \geq 10x - 60. Rearranging the terms, we get 120≥x120 \geq x or x≤120x \leq 120.

Therefore, all real numbers less than or equal to 120 are solutions to the inequality. The solution set is (−∞,120](-\infty, 120].

Exercise: EXERCISE 5.1 | Q: 12 | (Chapter: Page 7)
For More Understanding

Explanation

The question requires solving a linear inequality involving fractions. The provided context (specifically Example 6 and Example 4) demonstrates the standard procedure: simplify the inequality, multiply by the LCM of denominators to clear fractions, and then solve for the variable while adhering to Rule 2 regarding inequality signs. Since the multiplier (30) is positive, the inequality sign remains unchanged. The final solution matches the format of solution sets described in the examples (e.g., Example 4).

Solution Steps

  1. Step 1: Simplify the inequality: 12(3x5+4)≥13(x−6)  ⟹  3x10+2≥x3−2\frac{1}{2}\left(\frac{3x}{5} + 4\right) \geq \frac{1}{3}(x - 6) \implies \frac{3x}{10} + 2 \geq \frac{x}{3} - 2.

  2. Step 2: Multiply by the LCM (30): 30(3x10+2)≥30(x3−2)  ⟹  9x+60≥10x−6030\left(\frac{3x}{10} + 2\right) \geq 30\left(\frac{x}{3} - 2\right) \implies 9x + 60 \geq 10x - 60.

  3. Step 3: Solve for xx: 60+60≥10x−9x  ⟹  120≥x60 + 60 \geq 10x - 9x \implies 120 \geq x, which means x≤120x \leq 120.

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