Inequations Help?

Question

how do i do inequations with quadratics such as -

2/(2-x) Greater or equal to 3

using regions on a number line and testing them?

these questions have 2 answers. the answer to the one i gave as an eg is

4/3 less than or equal to x <2

oops. not necessarily 2 but you get the picture.

Answer 1

This doesn't look quadratic to me.

If 2-x > 0, then
2 >= 6-3x
0>=4-3x
3x>=4
x >= 4/3

If 2-x < 0, then the inequalities have to be reversed, giving x <= 4/3.
However, 2-x < 0 implies x > 2.
There are therefore no results from this option.

x >= 4/3 is the only answer.

Answer 2

a million/x > a million/(x + 2) x cannote equivalent 0 or -2 via fact this leads to branch with the aid of technique of 0. evaluate the circumstances x<-2, -20. a million/x > a million/(x + 2) If x>0 then multiply each and each and each and each little thing with the aid of technique of the effectual variety x(x+2). x + 2 > x it rather is consistently real. So the inequality is consistently real on a similar time as x>0. a million/x > a million/(x + 2) If -2 a million/(x + 2) If x<-2 then multiply with the aid of technique of the effectual variety x(x+2). x + 2 > x it rather is consistently real so the inequality is real for all x < -2. very final answer: x is below -2 or extra ideal than 0.

Answer 3

2 / (2 - x) ≥ 3
2 ≥ 3 (2 - x)
2 ≥ 6 - 3x
3x + 2 ≥ 6
3x ≥ 4
x ≥ 4/3 excluding x = 2
x= 2 is said to be a vertical asymptote.
Curve approaches asymptote but never meets it.

As x approaches line x = 2 from left, f(x)-> ∞
As x goes past line x = 2 (ie just to right of x = 2) ,
f (x)-> ∞
As x-> ∞ , y-> 0 (and is + ve)
As x-> - ∞ , y-> 0 (and is +ve)
ie the x axis is a horizontal asymptote
Curve cuts y axis at x= 0, y = 1 ie at (0 , 1)

Answer 4

2/(2-x)>= 3
2 >=3(2-x)
2>= 6-3x
3x>= 6 - 2
3x>=4
x >= 4/3 ..............(1)

Now,
Draw a line for the equation x = 4/3
this line passes through origine.

Now randomly select a point from this graph, say (1,1)
Sub. this point in equation (1),
you get,

1>= 4/3, which is not true.
So your region is the opposite side of this point including line means the region does not include point (1,1)

Answer 5

2/(2-x)>=3
2/(2-x) - 3>=0
2-6+3x/(2-x)>=0
3x-4/(2-x)>=0
now there r 2 cases i.e N^r & D^r are both +ive or both r -ive.
case1:both +ive therefore 4/3 <=x<2(as x=2 eq. is not defined as it reaches infinity therefore x=2 ix no the soln.)
case2:in both being -ive there is no solution as one is x<=4/3 & other x>=2 which doesn't make the equation +ive or equal to zero.

Answer 6

2/(2-x) >3

2> 6-3x

-4> -3x

4< 3x

x<4 /3

Answer 7

2/(2-x)>=3 |*(2-x)
2>=3(2-x)
2>=6-3x
-4>=-3x |*(-1)
4<=3x
x<=4/3

x is in the interval (-infinite;4/3]

Answer 8

If u come to me i teach u all inequations. Or u can mail me.OK