Solve the following inequations $\frac{2x+4}{x-1}≥5$
Answer & explanation
Correct answer: option 1
The correct answer is Option 1: (1, 3)
We have $\frac{2x+4}{x-1}≥5⇒\frac{2x+4}{x-1}-5≥0$
$⇒\frac{2x+4-5(x-1)}{x-1}≥0⇒\frac{2x+4-5x+5}{x-1}≥0$
$\frac{-3x+9}{x-1}≥0$ [Multiplying both sides by -1]
$\frac{3x-9}{x-1}≤0⇒\frac{3(x-3)}{x-1}≤0$ [Dividing both sides by 3]
$\frac{x-3}{x-1}≤0⇒1<x≤3⇒x∈(1,3]$