Question

Help please thanks :D
Linear inequalities: When do we have to consider cases (e.g. when x<1) and when do we not have to?

Coz I was just doing this question which apparently needs cases but I don't understand why.
2/(3(x+1)) + 1/(3(x+2))≤1/(x+1)

Answer 1

and also how do you solve the above inequality i thought i could but i can't apparently

Answer 2

the technique is to compare the expression to zero
it's always "safe" to add or subtract a quantity (but you can get into trouble if you multiply or divide). So write the problem as
\[ \frac{2}{3(x+1)} + \frac{1}{3(x+2)} - \frac{1}{x+1} \le 0 \]

Answer 3

yup then find the common denominator?

Answer 4

yes, combine the fractions

Answer 5

x+3/(x+1)(x+2)(x+1)

Answer 6

I get something else

Answer 7

which is?

Answer 8

what are you using for a common denominator ?

Answer 9

initially after combining it's something like (3x-9)/3(x+1)(x+2)(x-1) = x-3/(x+1)(x+2)(x-1)

Answer 10

I would use 3(x+1)(x+2) as the common denominator

Answer 11

ok

Answer 12

any particular reason?

Answer 13

if you do, multiply the first fraction by (x+2)/(x+2)

Answer 14

if 3(x+1)(x+2) as the common denominator what's the numerator?

Answer 15

the reason is that is the simplest common denominator.

***what's the numerator? ***
multiply the first fraction by (x+2)/(x+2)

Answer 16

as in x+3 or 3x-9?

Answer 17

I would never distribute in these problems. we want to keep things factored

the first fraction becomes
\[ \frac{2}{3(x+1)} \cdot \frac{(x+2)}{(x+2)} \]

Answer 18

isn't it 2/3(x-1) not x+1?

Answer 19

maybe you should post a screen shot? I am looking at what you posted in the question up top. If there are typo's , this will be very confusing...

Answer 20

erm it's hardcopy but i'll type it out \[\frac{ 2 }{ 3(x-1) }+\frac{ 1 }{ 3(x+2) }≤\frac{ 1 }{ x+1 }\]

Answer 21

oh, that is different from your question where it's (x+1) in the first fraction

Answer 22

OH CRAP SORRY

Answer 23

so your answer up above is mostly correct (except for x+1 twice and no x-1)
***x+3/(x+1)(x+2)(x+1)***
you mean
\[ \frac{(x+3)}{(x-1)(x+1)(x+2) }\]

Answer 24

so the problem is
\[ \frac{(x+3)}{(x-1)(x+1)(x+2) } \le 0 \]

Answer 25

ok yup :D

Answer 26

we need that fraction to be 0 or negative

Answer 27

it's zero for x= -3. make a note of that

Answer 28

that's where cases come in yay. looks like there'll be a lot?

Answer 29

yes, it looks a bit messy. I would plot a number line, and figure out the signs of each factor for each region of the number line.

Answer 30

ok so am i right to say that er the cases are something like ok wait. gimme a moment sorry

Answer 31

case 1: x+3≤0 and everything else ≥0
x>1

case 2: x-1≤0 and everything else more than 0
-1<x<1

case 3: x+2≤0 everhting else ≥0
-3<x<2 or x>1

case 4: x+1≤0
-3<x<-1 or x>1??

Answer 32

sorry forgot the equal sign for inequality

Answer 33

notice we can't let the bottom factors be 0 (no divide by 0)

Answer 34

ahh

Answer 35

case 4 should be x>-1 i think

Answer 36

graphically