Question

Simplify:
|1-a+b|-|a-b|
--------------
|b-a| +a

Answer 1

\[\frac{1-a+b-a+b}{b-a+a} = \frac{1-2a + 2b}{b}\]

all you've got to do is simplify by multiplying the sign to the values in brackets and compute ^_^

Answer 2

those aren't brackets :):):):)
absolute value :S

Answer 3

hmm , oh right, you're right lol. If you have absolute values, then the signs switch.

Answer 4

not sure how though? >_< I'm sorry. I know that |b-a| = a-b

Answer 5

but not sure of the one in the denominator.

Answer 6

i'm not sure how to solve this 2.....thnx a lot anyway :) ^_^

Answer 7

numerator*

Answer 8

Alright, let me see if I can make it simpler, |4x| = 4x, |-4x| = 4x

|4x-3|= 3-4x
|2a +b| = 2a+b

makes any sense?

Answer 9

yea it does...thnx :)so i just switch signs...nothing else?

Answer 10

looks like it :)

np dear ^_^

Answer 11

you switch signs only when you have a negative inside with another variable, others stay the same ^_^

Answer 12

hmm how abt -|a-b|???? is it a-b?

Answer 13

first find |a-b| = b-a then multiply it by (-) = -(b-a) = a-b LOL it's the same thing :)

Answer 14

so yes ^_^

Answer 15

^_^ Tjnx a lot :):):) ^_^

Answer 16

np dear :)

Answer 17

thnx*

Answer 18

For equations having terms with absolute value, you have to calculate twice, once if the value inside the brackets is positive (non-negative in a broader sense), and then if the value inside the brackets is negative. So, you have to think about cases, when (a-b)<0, when 0<=(a-b)<=1 and when (a-b)>1.

Now the solution. 1st case: when (a-b)<0 or a<b, |a-b| = b-a.
And, |1-a+b| = |1-(a-b)| = 1-a+b. |b-a| = b-a.
So, the simplification will be 1/b.

2nd case: when 0<=(a-b)<=1, then |a-b| = a-b, |1-a+b| = |1-(a-b)| = 1-a+b. |b-a| = a-b. So the simplification will be (1-2a+2b)/(2a-b).

3rd case: when (a-b)>1, then |a-b| = a-b, |1-a+b| = |1-(a-b)| = a-b-1, |b-a| = a-b. So the simplification will be 1/(b-2a).

Enjoy. :)

Answer 19

Oh thanks a lot :):):)