Question 36 of 40intermediate🔧 ApplyNumerical4 marks

Solve the inequalities and represent the solution graphically on number line: 5(2x−7)−3(2x+3)≤0,2x+19≤6x+475(2x - 7) - 3(2x + 3) \leq 0, \quad 2x + 19 \leq 6x + 47

Correct Answer

The solution set is x∈[−7,11]x \in [-7, 11] or −7≤x≤11-7 \leq x \leq 11.

Exercise: Miscellaneous Exercise on Chapter 5 | Q: 10 | (Page Page 10)
For More Understanding

Explanation

This question involves solving a system of two linear inequalities. Following the method shown in Example 11 of the textbook, we solve each inequality separately and then find the intersection of both solution sets. The first inequality gives us x≤11x \leq 11 and the second gives us x≥−7x \geq -7. The common solution is the interval where both conditions are satisfied simultaneously.

Solution Steps

  1. Step 1: Solve the first inequality 5(2x−7)−3(2x+3)≤05(2x - 7) - 3(2x + 3) \leq 0 Expanding: 10x−35−6x−9≤010x - 35 - 6x - 9 \leq 0 Simplifying: 4x−44≤04x - 44 \leq 0 Therefore: 4x≤444x \leq 44 or x≤11x \leq 11

  2. Step 2: Solve the second inequality 2x+19≤6x+472x + 19 \leq 6x + 47 Subtracting 2x2x from both sides: 19≤4x+4719 \leq 4x + 47 Subtracting 47: −28≤4x-28 \leq 4x Dividing by 4: −7≤x-7 \leq x or x≥−7x \geq -7

  3. Step 3: Find the common solution From inequality (1): x≤11x \leq 11 From inequality (2): x≥−7x \geq -7 The solution set is: −7≤x≤11-7 \leq x \leq 11 or x∈[−7,11]x \in [-7, 11]

  4. Step 4: Graphical representation on number line Draw a number line. Mark points at −7-7 and 1111. Draw a closed circle (filled dot) at x=−7x = -7 and x=11x = 11 (since both endpoints are included). Shade the region between −7-7 and 1111 to represent all values of xx satisfying both inequalities.

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