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If $\frac{3x-5}{6}+8≥4+\frac{2x}{3}$, then |
$x ∈ (-∞, 19]$ $x ∈ (-∞, -19]$ $x ∈ [19,∞)$ $x ∈ [-19,∞)$ |
$x ∈ (-∞, 19]$ |
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: $43-24 \ge 4x-3x$ $19 \ge x$ $x \le 19$ |
