Solving Equations with Negative CoefficientsDate: 12/13/2005 at 00:24:10 From: Alejandra Subject: negative coefficient When you solve an equation and the x or the variable is negative, like 8 - x = 3 do I multiply it by -1 so that it would be positive? Date: 12/13/2005 at 09:13:18 From: Doctor Peterson Subject: Re: negative coefficient Hi, Alejandra. There are two ways you can go here. One is, as you suggest, to multiply both sides by -1: 8 - x = 3 Subtract 8 from both sides -x = -5 Multiply both sides by -1 x = 5 (Actually, this is just the same as the usual last step, dividing both sides by the coefficient, which is -1.) The other approach (which helps even more with inequalities) is to avoid negative coefficients by moving the x to the other side: 8 - x = 3 Add x to both sides 8 = x + 3 Subtract 3 from both sides 5 = x Here I added x to both sides so that it would end up with a positive coefficient; then I isolated it on the right side. The result is the same both ways. You can do the same thing in more complicated cases like this: 8 - 5x = 2 - x Here, since the -5 on the left is less than the -1 coefficient on the right, I choose to get rid of the SMALLER multiple of x by adding 5x to both sides: 8 = 2 + 4x Now I have a positive coefficient, and the rest is easy: 6 = 4x 6/4 = x x = 3/2 If I kept x on the left, I would have to divide by a negative coefficient--which works fine, but gives me one more place where I can make a mistake: 8 - 5x = 2 - x 8 - 4x = 2 -4x = -6 x = -6/-4 = 3/2 It's the step of dividing by a negative, and the resulting sign change (in inequalities), that can be error-prone. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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