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

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embizze's profile pic

Posted on

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



oldnick's profile pic

Posted on


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

where did you read equal sign in my solution?


oldnick's profile pic

Posted on

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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embizze's profile pic

Posted on

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)`

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