Question

solving inequalities 1/(2x+1) + 1/(x+1) < 8/15

Answer 1

x >2 got it!

Answer 2

\[\frac{1}{2x+1}+\frac{1}{x+1}-\frac{8}{15}<0\]
\[\frac{(x+1)(15)+(2x+1)(15)-8(2x+1)(x+1)}{(2x+1)(x+1)(15)}<0\]

Answer 3

got it, but i was wrong!

Answer 4

\[\frac{15x+15+30x+15-8(2x^2+2x+x+1)}{15(2x+1)(x+1)}<0\]

Answer 5

this one really stinks. 4 terms, 5 intervals i will stop heckling and let myininaya complete it. have fun

Answer 6

\[\frac{-16x^2-24x+45x-8+30}{15(2x+1)(x+1)}<0\]

Answer 7

why does the 15 got to the top ? i thought all it needed was the bottem to be mutiplyied by 15 and top mutiplied by 8, i drew a graph like in class today, but i forgot how i got the intervals

Answer 8

\[\frac{-16x^2+21x+22}{15(2x+1)(x+1)}<0\]

Answer 9

i subtracted 8/15
then i found a common denominator

Answer 10

i subtracted 8/15 on both sides*

Answer 11

we need to factor the numerator
\[-16x^2+21x+22\]

Answer 12

=-16(22)=-11*(32)
21=32-11
so
21x=32x-11x
\[-16x^2+32x-11x+22=-16x(x-2)-11(x-2)=(x-2)(-16x-11)\]

Answer 13

so we have
\[\frac{(x-2)(-16x-11)}{15(2x+1)(x+1)}<0\]

Answer 14

so the top is zero when x-2=0 or -16x-11=0
the bottom is zero when 2x+1=0 or x+1=0

so we need to test around four numbers -1,-11/16,-1/2,2

Answer 15

yeah wow. see 4 zeros, 5 intervals to consider

Answer 16

glad you are doing this and not me

Answer 17

you want to finish it?

Answer 18

i will give you a medal

Answer 19

or can you finish it taz?

Answer 20

oh thankyou so much for helping me, a bit slow today : / and tutor is giving me the I am dissappointment look if I asked him ahaha

Answer 21

yes i know how to complete it from here thanks

Answer 22

cool i will help if you need more help though on it
it is a long problem
i just need a little break from it lol

Answer 23

so when it is postive is from - infinity to -0.5 , -11/16 to -1, and 2 to positive infinity

Answer 24

careful with your intervals here.

Answer 25

you have the right numbers, just in the wrong order.
- infinity< -1 < -11/16 < -.5 < 2

Answer 26

omgosh i noticed the quite the stupid mistake i have bad, i placed 1/2 infront of -1 :S

Answer 27

by which i mean you have the right numbers but you put them in the wrong order. it should be
\[(-\infty,-1)\cup (-\frac{11}{16},-\frac{1}{2})\cup (2, \infty)\]