Question

Solving inequality : attached pic is qn & ans, couldn't get the answer :(

Answer 1

wheres the pic

Answer 2

sorry new at this ><

Answer 3

alright \[ \frac{ x }{ x-4 }\le3\] to get rid of fractions we do reciprocals multiply x-4 to both sides \[x \le3x-12\] solve for x

Answer 4

can you do it from here or need help?

Answer 5

get x on one side of the sign so subtract 3x from x
\[-2x \ge-12\]

Answer 6

oops >.< \[-2x \le -12\] we flip the sign after dividing

Answer 7

x = 6

Answer 8

which way does the sign face?

Answer 9

oo !! didnt realise i could use this method, i used the sign test method >< I will try this method now ! Can u find out what went wrong in my sign test method ? sorry and thanks !!

Answer 10

not understanding what you did after\[\frac{ -2x-12 }{ x-4 }\]

Answer 11

oo my sch taught this method haha but if using yr method, after getting \[x \ge 6\], how do I link it to \[e ^{x}\] ?

Answer 12

cant see the whole problem cut off after 3

Answer 13

that is the full problem alr though o.o

Answer 14

theres no exponent on 3?

Answer 15

yep ! just 3 by itself

Answer 16

\[\frac{ e ^{x} }{ e ^{x}-4 }\le3?\]

Answer 17

yes !!

Answer 18

ok whats the answers?

Answer 19

\[x \ge6 \] or \[x <4\]

Answer 20

do the reciprocal of e^x/e^x-4 andmultiply both sides by e^x-4
\[e ^{x} \le 3e ^{x}-12\]

Answer 21

get e on one side so subtract 3 into e\[-2e ^{x}\le-12\]

Answer 22

divide by -2
\[e^x \ge6\]

Answer 23

sign flipped because you divided by a negative and 6 is positive because 2 negatives make a positive

Answer 24

@phi @jdoe0001 is this right or something need to be done with exponent on e?

Answer 25

but they want it in terms of x and not\[e ^{x}\] .. TT

Answer 26

i dont think it matters we got an answer and its in your choices @tejasvir ?

Answer 27

@SolomonZelman ?

Answer 28

looking at your work, it looks like you got a sign wrong


\[ \frac{x}{x-4} ≤ 3 \\ \frac{x}{x-4} -3 ≤ 0 \\ \frac{x}{x-4} - 3\frac{x-4}{x-4} ≤ 0 \\
\frac{x-3x+12}{x-4} ≤ 0 \\
\frac{-2x+12}{x-4} ≤ 0
\]

Answer 29

so the critical values are x=4 and x=+6 (not -6)

and the solution is x<4 or x≥ 6

Answer 30

oo !! thanks alot !

Answer 31

I noticed we can't let x=4 (it will cause a divide by 0) so x<4 (not x≤4)

Answer 32

ahh.. thank you !! but how do we link it to the \[e ^{x}\] then ?

Answer 33

the answer should be x≥ ln(6) or x< ln(4)

Answer 34

i thought so too, there must be a ln ..thank you so much !

Answer 35

replace x with e^x and then solve for x
x<4 or x≥ 6
becomes
e^x <4 so x < ln(4)
e^x≥ 6 so x ≥ ln(6)

Answer 36

If they don't show the ln, then you are the victim of a typo or oversight (happens a lot in math books)

Answer 37

haha yep this was my sch homework, thanks alot !! ^^