# x-3/x-5>0..... how is this inequality solved?

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Solve `(x-3)/(x-5)>0` :

A rational expression is positive if the signs of the numerator and denominator are the same; i.e. both positive or both negative.

x-3 is positive for x>3, and negative for x<3

x-5 is positive for x>5 and negative for x<5

x-3 and x-5 are both positive for x>5

x-3 and x-5 are both negative for x<3

**Thus the solution is x<3 or x>5**

FOR EMBIZZE:

Indeed my solution is : `-oo <x<3` and `5<x < +oo`

where did you read equal sign in my solution?

Ambigue description:

if inequality is: `(x-3)/(x-5)>0` we have to found points where ratio is positive, that is point where numerator and denominator have same sign (or both positive , or both negative)

Studing sign for `x-3 >0` that is `x>3`

and `x-5>0` means `x>5`

Let's draw graph about:

First line above is sign of ineqaulity: `x-3>0`

Second line( in the middle) si about ineqaulity `x-5>0`

(Red for positive values, black for negative ones)

Third (last below) is the sign results for ratio `(+):(+)= (-) :(-) = +` and `(+ ): (- )= (-) :(+)= -`

Now , let you see, in the intervall `(-oo; 3]` ratio is between two negative value, so positive, instead in the intervall `(3;5]` is a ratio between a positive value ( ineqaulity `x-3>0)` and a negative one ( `x-5>0)`

Instead in the intervall `(5; +oo)` are both positive, then ratio is positive too.

Then the third line below( ratio line) is to be drawn, as red until `(-oo; 3)` ,black in `[3;5)` and again red in `[5;+oo)`

So soltuions held: `-oo < x <3` ; `5< x<+oo` ` `

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But if x=3 the rational expression equals zero; this is not greater than zero thus the intervals should be `(-oo,3)uu(5,oo)`