Solve the given inequality for real\(x: \frac{1}{2}(\frac{3x}{5}+4) ≥ \frac{1}{3}(x-6)\).
Given Inequality:
\( \frac{1}{2}\left(\frac{3x}{5}+4\right) \ge \frac{1}{3}(x-6) \)
Step 1: Simplify both sides
Left-hand side:
\( \frac{1}{2}\left(\frac{3x}{5}+4\right) = \frac{3x}{10} + 2 \)
Right-hand side:
\( \frac{1}{3}(x-6) = \frac{x}{3} - 2 \)
Step 2: Form the inequality
\( \frac{3x}{10} + 2 \ge \frac{x}{3} - 2 \)
Step 3: Eliminate fractions
Multiply both sides by LCM of 10 and 3, i.e., 30:
\( 9x + 60 \ge 10x - 60 \)
Step 4: Solve the inequality
\( 9x - 10x \ge -60 - 60 \)
\( -x \ge -120 \)
Dividing by −1 (inequality sign reverses):
\( x \le 120 \)
Graphical Representation on Number Line:
A closed circle at 120 and the region to the left of 120 is shaded.
Final Answer:
The solution set is
{ x ∣ x ≤ 120 }