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Solving another Equation in One Variable

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!
Please I need some help! 


-Doug Amouzouvi

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 
carefully  multiplying out the brackets. You can either start with 
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 
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 [] 

              -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

Associated Topics:
Middle School Equations

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