Question

find set values of x for the following inequality

|4div(x-1)|>3-(3divx)

Answer 1

\[|4\div(x-1)|>3-(3\div x)\]

Answer 2

someone help...

Answer 3

hang-on... i need to go out for a while.... i will be back in half an hour..... bye...

Answer 4

Ok, I have worked on this for a while and are hitting many pitfalls. The idea is to split it into two cases, one where x-1 is positive and the absolute signs vanish and one where it is negative and the inequality reverses

Answer 5

\[3 - \frac{3}{x} = \frac{3x-3}{x}\]
2 cases
\[\frac{4}{x-1} > \frac{3x-3}{x}\]
and
\[\frac{4}{x-1} < \frac{-(3x-3)}{x}\]
Solve each case separately
I will look at the negative case first, cross-multiply
\[4x <-(3x-3)(x-1)\]
\[\rightarrow 3x^{2}-2x+3 < 0\]
You will find this has complex solutions so there are no x-values such that its less than 0
next case
cross-multiply
\[4x > (3x-3)(x-1)\]
\[\rightarrow 3x^{2}-10x+3 < 0\]
Factor
\[(3x-1)(x-3) < 0\]
For this product to be negative, one term must be positive, the other negative.
This occurs when x is between 1/3 and 3.
However, when x=1 the original equation becomes undefined
Solution\[x \in (1/3,1) U (1,3)\]

Answer 6

i skipped some of the algebra steps
do you see how i split the absolute value into 2 cases

Answer 7

ok... thx... i think i gonna to switch subject... coz tmr i hv chemistry exam... do u know someone good in chemistry?? coz now just 1 online... that's me...

Answer 8

shoot, no i don't sorry.
i took a couple years of it but i dunno if i can help you

Answer 9

ok.. nvr mind... i try to review my book.... no notes... ^^

Answer 10

good luck