Date: 07/14/98 at 10:09:08 From: James Subject: Transposing I am having trouble when transposing. For example, if y = 2x + 3, and I want to find the value of x, is it -3 + y = 2x + 3 - 3 or 3 - y = 2x + 3 - 3 ? I understand how to transpose the 2 by division, but when it comes to equations like these I get confused. Also, is there any rule to which order I can transpose a more complex equation, such as y = 2x + 3(2x^2 + 5) ?
Date: 07/22/98 at 11:12:39 From: Doctor Rick Subject: Re: Transposing Hi, James, I think what will help you is to think about why transposing works. The basic idea is that if you have two things that are equal, then you can add the same thing to both and the results will be equal. In symbols, if a = b then a + c = b + c. The same holds for subtracting, multiplying, or dividing (or raising to a power, etc.): a-c = b-c a*c = b*c a/b = b/c a^c = b^c Now, can you see how it works? If y = 2*x + 3, we can subtract 3 from both sides to get y - 3 = 2*x + 3 - 3. If you subtract 3 from the right side but subtract the left side from 3, you aren't doing the same thing to both sides. When you have a more complicated equation, just break it down in the reverse order from the way it was built up. Before we begin, I will add parentheses and operation symbols to your equation to make all the operations and their order explicit. The equation is: y = 2 + (3 * ((2 * (x^2)) + 5)) To build the right side, first you squared x, then you multiplied it by 2, added 5, and multiplied the result by 3. Then you added 2. So in transposing, you reverse the process. Read down the left column, and up the right. Building Transposing -------- ----------- x^2 | ^ sqrt((((y-2)/3)-5)/2) = x (take square root) 2*(x^2) | | (((y-2)/3)-5)/2 = x^2 (divide by 2) (2*(x^2))+5 | | ((y-2)/3)-5 = 2*(x^2) (subtract 5) 3*((2*(x^2))+5) | | (y-2)/3 = (2*(x^2))+5 (divide by 3) 2+(3*((2*(x^2))+5) v | y-2 = 3*((2*(x^2))+5) (subtract 2) And there's your answer: x = sqrt(((y-2)/3 - 5)/2). If you like a slogan to help you remember the order, it could be "last on, first off," or "from the outside in." I hope this helps. If you need more help, just ask. - Doctor Rick, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum