What is (x-2}{x-4}>= \frac{x+2)/x ?

The solution to (x-2}{x-4}>= \frac{x+2)/x is x 4

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(x-2}{x-4}>= \frac{x+2)/x

\frac{x - 2}{x - 4} \geq \frac{x + 2}{x}

x < 0 or x > 4

Interval Notation

(- \infty, 0) \cup (4, \infty)

Solution steps

\frac{x - 2}{x - 4} \geq \frac{x + 2}{x}

Rewrite in standard form

\frac{1}{x ( x - 4 )} \geq 0

\frac{1}{a} \geq 0

then

a > 0

x (x - 4) > 0

Identify the intervals

x < 0 or x > 4

4 j^{2} = - 32 + 66 j

∣ 2 x + 1 ∣ \geq 2 + x

5 y + (- 5) = 10

y^{2} - 10 y + 26 = 0

x (x - 1) - 5 (x - 2) = 2

What is (x-2}{x-4}>= \frac{x+2)/x ?
The solution to (x-2}{x-4}>= \frac{x+2)/x is x<0\lor x>4