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### Solving another Equation in One Variable

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Date: 1/12/96 at 20:39:18
From: Anonymous
Subject: Solving Equations

Hi! Dr.Math, I am having some trouble with some Math Equations!
I'm  trying to learn this from this from my math teacher, but she
seems to  talk so fast!  I have asked for some of your help before
and you have  given me success on my test!  Could you please go
over the question I wrote down?

-2[3-2(x-1) + 5(2-x)] = -3[x+2(x-1)]

I tried out this question and I don't know what step to take first!

Thanks

-Doug Amouzouvi
```

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Date: 6/5/96 at 8:18:33
From: Doctor Charles
Subject: Re: Solving Equations

The thing to do with this sort of question is to go slowly and
the inner  brackets and work out or the outer brackets and work in.
I like to work  from the inside out.

-2[3-2(x-1) + 5(2-x)] = -3[x+2(x-1)]

so first the () brackets, remembering to multiply out every term in
the
bracket.     -2(x-1) = -2x + 2
+5(2-x) = 10 - 5x
+2(x-1) = 2x - 2

so            -2[3-2x+2+10-5x]  = -3[x+2x-2]

now add the x terms and the constant terms together in each []
bracket

-2[15 - 7x] = -3[3x - 2]

Multiply out again

-30 + 14x  = -9x + 6

Move x terms to the left, constants to the right:

23x  = 36

so                     x  = 36/23  = 1 13/23.

-Doctor Charles,  The Math Forum

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Associated Topics:
Middle School Equations

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