What is 1/(x-3)+3/(x-7)>= 0 ?

The solution to 1/(x-3)+3/(x-7)>= 0 is 3 7

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1/(x-3)+3/(x-7)>= 0

\frac{1}{x - 3} + \frac{3}{x - 7} \geq 0

3 < x \leq 4 or x > 7

Interval Notation

(3, 4] \cup (7, \infty)

Solution steps

\frac{1}{x - 3} + \frac{3}{x - 7} \geq 0

Rewrite in standard form

\frac{4 x - 16}{( x - 3 ) ( x - 7 )} \geq 0

Factor

\frac{4 x - 16}{( x - 3 ) ( x - 7 )} : \frac{4 ( x - 4 )}{( x - 3 ) ( x - 7 )}

\frac{4 ( x - 4 )}{( x - 3 ) ( x - 7 )} \geq 0

Divide both sides by

4

\frac{\frac{4 ( x - 4 )}{( x - 3 ) ( x - 7 )}}{4} \geq \frac{0}{4}

Simplify

\frac{x - 4}{( x - 3 ) ( x - 7 )} \geq 0

Identify the intervals

3 < x \leq 4 or x > 7

∣ 2 x - 3 ∣ = 19

\frac{2}{7} y = - 10

16 x^{2} - 32 x + 16 = 0

5125

\frac{7}{2} \cdot 3

What is 1/(x-3)+3/(x-7)>= 0 ?
The solution to 1/(x-3)+3/(x-7)>= 0 is 3<x<= 4\lor x>7