inequalities with x in the denominator (1 Viewer)

[kr73114]

i got down to this stage...
(2x-3)(x-2)>(x-2)^2
=(x-2)^2-(2x-3)(x-2)
i know i have to take (x-2) as common...how do i work it out?

 

[SeCKSiiMiNh]

your question isnt clear

what is the original question?

 

[kr73114]

soz i hope this is better
(2x-3)(x-2)>(x-2)^2

(x-2)^2-(2x-3)(x-2)>0

...now what next
um the original question is:
2x-3/x-2>1

 

[AmieLea]

(2x-3)/(x-2)>1

(2x-3)(x-2)>(x-2)^2

0>(x-2)^2-(2x-3)(x-2)

0>(x-2) [(x-2)-(2x-3)]

0>(x-2)(x-2-2x+3)

0>(x-2)(1-x)

(x-2)(1-x)<0

(solve graphically)

and x<2; x>1

1<x<2

gotta love these questions

 

[Rezen]

Solving graphically yields x<1, x>2.

There is also the case method.
for (2x-3)/(x-2)>1 either x<2 or x>2

case 1; when x<2,

(2x-3)/(x-2)>1

2x-3 < x-2

x<1

case 2; when x>2

(2x-3)/(x-2)>1

2x-3>x-2

x>1, but x>2

therefore solution to this inequality is x>2

it follows solutions to (2x-3)/(x-2)>1 are x<1, x>2.

 

[ohexploitable]

Take (x-2) as a common factor then graph it.

 

[ohexploitable]

x<1 or x>2

 

[Affinity]

you don't really need to multiply through by the denominator..
(2x-3)/(x-2)>1
(2x-4 + 1)/(x-2)>1
2 + [1/(x-2)]>1
1/(x-2) > -1
==> x-2 > 0 or x-2 < -1
x > 2 or x < 1