Solving Equations for VariablesDate: 02/03/2002 at 19:13:27 From: Anonymous Subject: Solving formulas for variables Hello! How do you solve the following equation for (M) and then (m)? k = M-m/1+Mm I tried to solve this, but I all I got was k+kMm+m = M. The real answer is M = m+k/1-km; m = M-k/1+km. Please show me how you get there, step by step. Thank you in advance. Date: 02/04/2002 at 02:07:28 From: Doctor Jeremiah Subject: Re: Solving formulas for variables Hi there! First we need to change the question. Here is what the question really is: k = (M-m)/(1+Mm) Notice the extra parentheses. Without them the question would be determined as: k = M - m/1 + Mm Which is: k = M - m + Mm And that isn't what you are expecting. So with the parentheses where they belong: k = (M-m)/(1+Mm) The first thing to do is to get rid of the parentheses, but the division sign is in the way, so first let's get rid of the division. Move the denominator over to the left-hand side of the equation. Since the right side is divided by the denominator, we can move it to the other side by multiplying both sides by the denominator: k = (M-m)/(1+Mm) (1+Mm)k = (1+Mm)(M-m)/(1+Mm) This works because now the right-hand side has (1+Mm) on both the top and bottom of the division. And (1+Mm) divided by (1+Mm) is 1 so: k = (M-m)/(1+Mm) (1+Mm)k = (1+Mm)(M-m)/(1+Mm) (1+Mm)k = (M-m) x (1+Mm)/(1+Mm) (1+Mm)k = (M-m) x 1 (1+Mm)k = (M-m) (1+Mm)k = M - m Okay, so we got rid of one set of parentheses. Now we need to get rid of the other set. There is a k multiplied by the parenthesis so we need to do something about that. If we "distribute" that k into the parenthesis, then we can get rid of the parenthesis. To "distribute" the k we need to multiply all the terms inside the parentheses by k: (1+Mm)k = M - m (1k + Mmk) = M - m 1k + Mmk = M - m k + Mmk = M - m Now we don't have any parenthesis. Now we have to decide what to solve for. I will choose M. That means we must get all the Ms on one side of the equals sign (by themselves) with everything else on the other side. That means moving the term that has the M on the left side of the equals sign over to the right side of the equals sign. Since that term is added to the left side, one way to move it to the right side is to subtract it from both sides of the equation: k + Mmk = M - m k + Mmk - Mmk = M - m - Mmk This works because Mmk - Mmk is 0 and anything plus 0 is unchanged: k + Mmk = M - m k + Mmk - Mmk = M - m - Mmk k + 0 = M - m - Mmk k = M - m - Mmk Now we have all the Ms on one side of the equals sign, but they need to be by themselves, so we need to get rid of the m term that is on the right side. One way to get it off the right side would be to move it to the left side. How? Since it is subtracted from the right side we can move it by adding m to both sides: k = M - m - Mmk k + m = M - m - Mmk + m Now just rearrange a little: k = M - m - Mmk k + m = M - m - Mmk + m k + m = M + m - m - Mmk Now m - m is 0 and anything added to 0 is unchanged: k = M - m - Mmk k + m = M - m - Mmk + m k + m = M + m - m - Mmk k + m = M + 0 - Mmk k + m = M - Mmk Now all the Ms are on the right side, by themselves. That's good. How do we combine terms to get just one M? Well, remember when we distributed earlier. What happens if we un-distribute? M(1-mk) is the same as M - Mmk so: k + m = M - Mmk k + m = M(1 - mk) We know it is the same because if we distribute the M into the parenthesis we get M - Mmk right back. Now we need the M by itself. That means moving the (1 - mk) term. Since it's multiplied onto the right side, we can move it to the left side by dividing both sides by (1 - mk): k + m = M(1 - mk) (k + m)/(1 - mk) = M(1 - mk)/(1 - mk) This works because (1 - mk) divided by (1 - mk) is 1, and anything times 1 is unchanged: k + m = M(1 - mk) (k + m)/(1 - mk) = M(1 - mk)/(1 - mk) (k + m)/(1 - mk) = M x 1 (k + m)/(1 - mk) = M If we start back where I said we had a decision to make and choose m instead, we can do the same sort of thing and end up with: m = (M - k)/(Mk + 1) Try it, and if you get stuck please write back. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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