Question

Linear inequalities

Answer 1

((x+a)/b)≥a/(x+b), where a,b>0

Answer 2

the answer is supposed to be -a-b≤x<-b
i can get -a-b≤x but I don't know how to get the other part

Answer 3

Have you assumed that x+b > 0?

Answer 4

no. technically x could be negative?

Answer 5

I would write your problem as
\[ \frac{x+a}{b}-\frac{a}{x+b} \ge 0 \]
or
\[ \frac{(x+a)(x+b) -ab}{b(x+b) }\ge 0 \]

Answer 6

hm...i cross multiplied and simplified to get
x^2+bx+ax+ab≥ab
= x^2+bx+ax≥0
=x(x+b+a)≥0
=x≥-b-a

Answer 7

simplify that to
\[ \frac{x(x+(a+b))}{b(x+b)}\ge 0 \]
b is >=0 so the "b" in the denominator it won't change the sign of that expression, so just consider
\[ \frac{x(x+(a+b))}{(x+b)}\ge 0 \]

Answer 8

...and when you did that, you assumed x+b > 0.
Now, go back and make the other assumption.

Answer 9

you have 3 factors, and if all 3 are positive, or 2 of the 3 are negative, you will get a positive result i.e. >=0

Answer 10

OH is it because if i assumed that x+b<0, then my answer would be
-a-b≥x instead of -a-b≤x?

Answer 11

assuming all 3 factors are >=0 gives your first result.
now consider x and x+ b both negative (luckily we know that is the only combo we have to consider), because the numbers must be x <= x+b <= x+a+b

Answer 12

sorry just to confirm what are the 3 factors?

Answer 13

\[ \frac{x(x+(a+b))}{(x+b)}\ge 0 \]

that expression is positive if x<0 & x+a <0 and (x+a+b)>0

Answer 14

**that expression is positive if x<0 & x+b <0 and (x+a+b)>0

Answer 15

first of all does that mean that this <0 and >0 can only be this way? like e.g. x cant be >0
secondly, does that mean that i'm wrong to say that x(x+b+a)≥0 since x<0?

Answer 16

I don't understand the first question.

for the second question
for x<0
x * (x+ a+b) will be negative for when x > -(a+b) because you will have
neg * plus = neg result
if x is less than -(a+b), then you would get neg*neg = positive result.

Answer 17

ah alright. But then how would I get x<-b?

Answer 18

let's back up a bit
if you know A<B<C
and you have the expression
A*B*C >= 0

it's clear if A is positive, then so is B and C (i.e. all >0) and their product is positive i.e >0
if A is neg and both B and C are positive, their product is negative (i.e. < 0) so we don't want that case

if A and B are neg and C is positive that has a positive product i.e. >0

does that all make sense ?

Answer 19

and we don't want all 3 negative, because their product is neg and we want the product >0

Answer 20

yup

Answer 21

now the three factors are
A= x
B= (x+b) (because we only care about the sign, it does not matter if we divide or multiply)
C= (x+a+b)

Answer 22

ok

Answer 23

so one case is
x<0 and x+b < 0 (at the same time, i.e. both are negative)
that becomes x<0 and at the same time x<-b
so x <-b is required for both quantities to be negative
(I think we don't want x<= -b because x=-b causes a divide by 0)

at the same time , we require the largest expression C = (x+a+b) >=0
which means x>= -(a+b)

Answer 24

YUP

Answer 25

u've been explaining this to me for about 20 minutes. thank you so much :D (wish i could fan u twice)

Answer 26

to be complete we have to consider the case where all 3 A,B and C are >=0
i.e.
x>=0
x>= -b
x>= -(a+b)
which boils down to x>=0

Answer 27

OH I THINK I GET IT. I SEE THE LIGHT. THANK YOU!!!!!!!