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$7≤3x+\frac{11}{2}≤11$ |
$[2,\frac{13}{2}]$ $[2,\frac{10}{3}]$ $[1,\frac{11}{2}]$ $[1,\frac{11}{3}]$ |
$[1,\frac{11}{3}]$ |
The correct answer is option (4) : $[1,\frac{11}{3}]$ $7≤3x+\frac{11}{2}≤11$ $14≤3x+11≤22$ $14-11≤3x≤22-11$ $3≤3x≤11$ $1≤x≤\frac{11}{3}$ ∴ Solution set is { x : x ∈ R, 1 ≤ x ≤ $\frac{11}{3}$ } i.e $[1,\frac{11}{3}]$ |
