easy question (that i can't solve) (1 Viewer)

[thisser]

i need your help with this question, which i thought looked really easy, but for some reason i can't get the answer. my brain has rusted after the holidays, lol

umm..

solve the following inequation

x + 2 / 1 - x < 1

please help.

 

[thisser]

actually, i've made progress (silly me) but there's a step I don't understand.

midway in solving it, i get

2x + 1 / 1 - x < 0

but how do I get to

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

?

 

[Winston]

solve for x?

 

[RyddeckerSMP]

x+2/1-x -1(1-x)/1-x <0
= 2x+1/1-x<0
=(1-x)x 2x+1/1-x <0 x 1-x
= (2x+1)(1-x)<0
then do your method for solving x
my method:
x= 1/2, 1
graph....
x<1/2, x>1

 

[C'est la vie]

That's right. *thumbs up*
It's kind of hard to follow through the working on the computer, but it's all good

Is anyone working from the yellow Fitzpatrick book at the moment? I'm doing the last excercises for sequences and series, and they kill me!!

Love Katie xox

 

[SamD]

RyddeckerSMP: should have been -1/2 not 1/2

An alternate solution...

(x+2)/(1-x)<1

Firstly x<>1 Can't have 0 denominator (<> means does not equal)

Now when the denominator is +ve then

1-x>1
x<1 ....(1)

And...

x+2<1-x
x<-1/2 ...(2)

(1) and (2) are both true only when x<-1/2 ...(3)
(This is definitely part of the solution).

Now if the denominator in the original is negative then,
1-x<0
x>1 ...(4)

And...

x+2>1-x (reverse the inequality sign as we're multiplying by a negative number)
x>-1/2 ...(5)

(4) and (5) are both true only when x>1 ...(6)

Now combine (3) and (6) to get the final solution....
x<-1/2 OR x>1

HTH
Sam

 

[shkspeare]

(x+2)(1-x) < (1-x)^2
(1-x)[x+2 - 1 + x ) < 0
(1-x)(2x+1) < 0
x > 1, x < -0.5, x <> 1

square ur denominator... look how easy it is... 4 lines

 

[SamD]

Yeah great, but I'd suggest a little more explanation is in order, this is not an Ext1 or 2 forum!

(x+2)/(1-x)<1

Firstly x<>1 (to avoid division by 0).

Multiply by the square of the denominator as this ensures we are multiplying by a positive number and hence the inequality sign is not affected. Furthermore it will allow canceling of the original denominator.

Hence...
(x+2)(1-x)<(1-x)^2
(x+2)(1-x)-(1-x)^2 <0

Now take (1-x) out as a common factor...
(1-x)[(x+2)-(1-x)] <0
(1-x)(2x+1)<0

The LHS equals 0 when x=1 or x=-1/2

Sub in some x value between -1/2 and 1, say 0
This doesn't work so the solution is outside these values
x<-1/2 OR x>1

HTH
Sam

 

[RyddeckerSMP]

yeah..... i think that my way is easier, way less confusing... but yeah x should equal -1/2... i though i stuffed it up somewhere