SOLUTION: 23-2[4+3(x-1)]+5[x-2(x+3)]=7 {x-2[5-(2x+3)] } this all looks like jumble to me, my I please get this problem broken down step by step; PLEASE

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: 23-2[4+3(x-1)]+5[x-2(x+3)]=7 {x-2[5-(2x+3)] } this all looks like jumble to me, my I please get this problem broken down step by step; PLEASE      Log On


   



Question 270585: 23-2[4+3(x-1)]+5[x-2(x+3)]=7 {x-2[5-(2x+3)] }
this all looks like jumble to me, my I please get this problem broken down step by step; PLEASE

Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
23-2[4+3(x-1)]+5[x-2(x+3)] = 7 {x-2[5-(2x+3)]
------------
23-2[4+3x-3]+5[x-2x-6] = 7 {x-2[5-2x-3]
----------------
23-2[1+3x]+5[-x-6] = 7{x-2[2-2x]
----------------
23-2-6x-5x-30 = 7{x-4+4x]
---------------
-9-11x = 7{5x-4]
------------------
-9-11x = 35x-28
46x = 19
------
x = 19/46
==============
Cheers,
Stan H.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!

I hope I copied it OK. I slightly changed it to
make it more readable to me.
You're right. It looks like a jumble to me, too, and
I'm pretty good with these things. What you have to do
to make it less of a jumble is to not look at the whole thing
at once, (like not looking down if you're climbing up a cliff)
-----------------------------------------------------
(a) start by identifying the innermost brackets only.
then pick just one of them and try to simplify it.
I'll start with the (2x - 3) at the end, since it is nested
inside 3 sets of brackets like this: ((( ))).
It's got a (-) in front of it, so it becomes
23-2%2A%284%2B3%2A%28x-1%29%29%2B5%2A%28x-2%2A%28x%2B3%29%29+=+7%2A%28x-2%2A%285-2x-3%29%29
I got rid of the innermost ().
(b) Now I'll move on to the next set of innermost ()s
which is the (x+3) in the middle inside 2 brackets:(())
It gets multiplied by the -2 in front, so
23-2%2A%284%2B3%2A%28x-1%29%29%2B5%2A%28x-2x%2B6%29+=+7%2A%28x-2%2A%285-2x-3%29%29
And I got rid of another set of ()s
Another innermost () is (x-3) moving left. It gets multiplied by
the 3 in front:
23-2%2A%284%2B3x-3%29%2B5%2A%28x-2x%2B6%29+=+7%2A%28x-2%2A%285-2x-3%29%29
That's 3 sets of ()s I eliminated
(c) Now look for more innermost ()s.
The (5 - 2x - 3) is one. It gets multiplied by the -2 in front, so
23-2%2A%284%2B3x-3%29%2B5%2A%28x-2x%2B6%29+=+7%2A%28x+-+10+%2B+4x+%2B+6%29
(d) Now there are no more innermost ()s so just choose among
the brackets that are left. I'll just do them all in one step:
23+-+8+-+6x+%2B+6+%2B+5x+-+10x+%2B+30+=+7x+-+70+%2B+28x+%2B+42
(e) Now just collect like terms:
23+-+8+%2B+6+%2B+30+-+6x+%2B+5x+-+10x+=+-+70+%2B+42+%2B+7x+%2B+28x
51+-+11x+=+-28+%2B+35x
35x+%2B+11x+=+51+%2B+28
46x+=+79
x+=+79%2F46
This could be wrong, since it's so easy to make a mistake
along the way, but the idea is what's important. Start from
the inside ()s and work your way out.