If $\frac{3x-5}{6}+8≥4+\frac{2x}{3}$, then
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $x ∈ (-∞, 19]$ **
Solve the inequality:
$\frac{3x-5}{6}+8 \ge 4+\frac{2x}{3}$
Multiply both sides by $6$:
$3x-5+48 \ge 24+4x$
$3x+43 \ge 24+4x$
Shift terms:
$43-24 \ge 4x-3x$
$19 \ge x$
$x \le 19$