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Review

anonymous · Answer

$$|2x-1| - |x+5| = 3 $$

solve for x

anonymous · Answer

@primeralph

shamil98 · Answer

Your equations are
-(2x-1) - (-(x+5)) = 3
and 
2x -1 -( x+5) = 3
distribute the negatives and solve for x

anonymous · Answer

@shamil98 could It ever be possible to evaluate it like 

(2x-1) - [-(x+5)] = 3 ?

anonymous · Answer

or [-(2x-1)]-(x+5) = 3 if not why?

shamil98 · Answer

You have two absolute equations
so no you can't
anything inside an absolute value symbol comes out positive which is why you have two solutions for an absolute value equation
for example
|x + 5| = 3
its simply
-(x+5) =3
or
x + 5 = 3

shamil98 · Answer

like if |x| = 2
x = 2, or -2

anonymous · Answer

alright. how about an inequality like 

$$|x-1|-|x-3| \ge 5 $$

shamil98 · Answer

-(x-1)-(-(x-3) >= 5
or
-x + 1 + x - 3  >= 5

shamil98 · Answer

$$- \times - = +$$

anonymous · Answer

so x = 3 and -3 for that answer... and for this 1 it is
it seems like the x cancel?

shamil98 · Answer

but in this case that inequality is false
-2 is not greater than or equal t o5.

shamil98 · Answer

the -x and x become zero

anonymous · Answer

so I would say this is a false ... Could I square both the terms and the other side and solve for x as a quadratic?

shamil98 · Answer

you would still end up with no solution

anonymous · Answer

alright. would it work with the equation above?

shamil98 · Answer

-(2x-1) - (-(x+5)) = 3
-2x + 1 + x + 5 = 3
-x + 6 = 3
-x + 3 = 0
a = 0
b = -1 
and c = 3
if everything is divided by 2(a)
or basically 0
it would not work
a >= 1
anything divided by zero is undefined

shamil98 · Answer

sorry, OS was updating again..

primeralph · Answer

Still need help?

anonymous · Answer

thanks guys