Question

solve for x here 6+2x+(3x^2)/(1-x)<3/(1-x)

Answer 1

\[\large \frac{(6+2x+(3x^2)}{(1-x)}<\frac{3}{(1-x)}\]
or
\[\large 6+2x+\frac{(3x^2)}{(1-x)}<\frac{3}{(1-x)}\]
...which?

Answer 2

the bottom one

Answer 3

Your answer is \[-1<x<3\]

Answer 4

the answer is x>3 but i dont understand why... when i do it i get x<3

Answer 5

Sorry, I think I made a mistake.
It should be \[1 < x <3\]

I don't see why it should be x>3.

Answer 6

i dont either... which is why im confused lol

Answer 7

Hmmm...

Answer 8

6+2x+((3x^2))/(1-x) < 3/(1-x)
6(1-x)+2x(1-x)+(3x^2) < 3
6-6x + 2x-2x^2 + 3x^2 < 3
6 - 6x +2x -2x^2 + 3x^2 < 3
6 - 4x + x^2 < 3
3 - 4x + x^2 < 0

solve the quadratic
3 - 4x + x^2 = 0
(x-3)(x-1) = 0

therefore solutions are:
x = 3 or x = 1

sub back in and test solutions:
if x = 1:
6+2x+((3x^2))/(1-x) < 3/(1-x)
6+2+((3))/(1-1) < 3/(1-1)
11/(0) < 3/(0)
cant divide by zero, so 1 isn't a solution

now test other solution:
if x = 3
6+2x+((3x^2))/(1-x) < 3/(1-x)
6+6+((27))/(1-3) < 3/(1-3)
39/-2 < 3/-2
-18.5 < -1.5

true, so 3 is the smallest solution

x>3

Answer 9

sorry, i type really slowly, cheers for waiting

Answer 10

i guess if u want to "prove" the maths, u can test any number smaller than 3 and any number larger than 3... it should hold up

Answer 11

Hurray!

Answer 12

thank you @jack1