Solving another Equation in One VariableDate: 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! Thanks -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 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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