Question

Solve the nonlinear inequality 3+x/3-x greater or equal 1

Answer 1

Help

Answer 2

\[\frac{ 3+x }{ 3-x }\ge1\]
multiply both sides by this
\[(3-x)^2\]

Answer 3

Why is it squared

Answer 4

Because you don't know what x is.

Answer 5

its not, I think he typed that by accident

Answer 6

it could be 9 and then you just ,mulktiplyiing both sides by a negative.

Answer 7

multiplying.

Answer 8

no....@sweet1137

Answer 9

Kk

Answer 10

no no that's not correct

Answer 11

I'm confused

Answer 12

Here's an example.

Answer 13

hold on, both of you

Answer 14

if x is 9 in your inequality.

Answer 15

\[\frac{ 3+9 }{ 3-9 }\ge 1\]
And you multiply by 3-9 to both sides.

Answer 16

\[\frac{ 12 }{ -6 }\times -6 \ge 1\times -6\]

Answer 17

you must change the inequalitysign.

Answer 18

that solves the inequality, x cannot be 3, since for x=3 the expression is undefined
yet for all other values of x greater than 0, the expression will yield values larger than 1

Answer 19

you're right

Answer 20

no...you must multiply by the square of the denominator so the inequality sign won't get affected...

Answer 21

my b, my b

Answer 22

yes, you're right

Answer 23

So what do I do??.

Answer 24

\[\frac{ (x+3)(x-3)^2 }{ x-3 }\ge (x-3)^2\]
\[(x+3)(x-3)\ge (x-3)^2\]
\[(x+3)(x-3)-(x-3)^2\ge 0\]
\[(x-3)[x+3-(x-3)]\ge 0\]
\[6(x-3)\ge 0\]

Answer 25

that still doesn't work, since x=0 is a solution not included in that inequality

Answer 26

Jeez

Answer 27

\[x-3\ge 0\]

Answer 28

SInce the solution would be x>3 since x cannot be 3

Answer 29

If x needs to be greater than 3, like your inequality states, why is it that for x = 0, 3+0/3-0 is greater than or equal to 1

Answer 30

But then you're right. I'm trying think back to my early days of doing this stuff.

Answer 31

hahah yea man, I'm trying too

Answer 32

Made a ,mistake

Answer 33

\[(3-x)(3+x-3+x)\ge 0\]

Answer 34

two solutions.
\[2x(3-x)\ge 0\]

Answer 35

Solved.

Answer 36

I did a boo boo.

Answer 37

How did u do that lmao