Question

Solve for x:
[(x-3)/(x-1)] less tahn or equal to [(4/(x+8)]

Answer 1

ick

Answer 2

last one for the night. you have to actually put everything on one side of the equal sign in one term

Answer 3

\[\frac{x-3}{x-1}\leq \frac{4}{x+8}\]
\[\frac{x-3}{x-1}-\frac{4}{x+8}\leq 0\]
\[\frac{(x-3)(x+8)-4(x-1)}{(x-1)(x+8)}\]
see why i said ick?

Answer 4

\[\frac{x^2-x+20}{(x-1)(x+8)}\leq 0\]
\[\frac{(x-5)(x+4)}{(x-1)(x+8)}\leq 0\]

Answer 5

still there?

Answer 6

yeah

Answer 7

zeros of these factors are -8, -4, 1, 5
so we have to break up the line into 5 pieces

----------- -8 ------------- -5 -------------- 1------------ 5 ---------------
and determine the sign on each interval

Answer 8

oh is this liek a number line?

Answer 9

i give you a hint. it will be positive , negative, positive, negative positive

Answer 10

yeah a number line broken into 5 parts at the zeros of each factor

Answer 11

it changes sign at each zero, so you really only have to check one and you will know them all

Answer 12

im kinda confused

Answer 13

from
\[(-\infty,-8)\] positive
from
\[(-8,-5)\] negative and so on

Answer 14

yeah am sure you are

Answer 15

it is all algebra up to this step
\[\frac{(x-5)(x+4)}{(x-1)(x+8)}\leq 0\]

Answer 16

painful annoying but necessary algebra. now you have 4 factors rigth?

Answer 17

yeah i se taht...so tehn u set tehm to 0 n u got -8, -4, 1, 5. I got it up to there

Answer 18

ok so you understand that
\[<0\] is a synonym for negative

Answer 19

so now the question is, for which values of x will this monstrosity be negative

Answer 20

it will change sign at those zeros. it will go from being positive to negative or negative to positive. so your job now is to figure out over which of those intervals it is positive, over which it is negative, and pick the negative ones as our answer

Answer 21

the intervals are
\[(-\infty,-8),(-8,-5),(-5,1),(1,5),(5,\infty)\]

Answer 22

and you have to pick the right ones. the ones for which it is negative. do you know how to tell?

Answer 23

sorry not really..teh comp frozwe

Answer 24

ok you can just check over any interval. pick any number. i pick 0.
plug in 0 and see what you get

Answer 25

you dont even have to compute

Answer 26

if you replace x by 0, two factors are negative and two are positive, so it will be positive

Answer 27

that means on the entire interval (-5,1) it is positive because 0 is in that interval. so it goes like this
Positive Negative Positive Negative Positive respectively

Answer 28

you want the negative ones, so your solution is
\[(-8,-5]\cup (1,5]\]

Answer 29

notice the careful placement of ( vs [
you have less than or equal to so you use [ or ] except if that number would make the DENOMINATOR 0. that is why i wrote
\[(-8,-5]\] open and closed

Answer 30

i see, wow thsi was a long and tedious probulem huh?

Answer 31

wait so is (−8,−5]∪(1,5] teh solution or (−8,−5]

Answer 32

both

Answer 33

oh ok tahnks soo much...i was wrking on ths ifor awhile n i cudnt figure it out