Substitution MethodDate: 01/08/2002 at 16:21:31 From: Erica Subject: The Substitution Method Please help me so I can better understand the substitution method. Here is the problem: 2x + y = 3 3x + 2y = 5 Thank you for your time, Erica Date: 01/08/2002 at 18:26:44 From: Doctor Ian Subject: Re: The Substitution Method Hi Erica, Suppose we have an equation like y = (something in terms of x) Then we can take one of your equations and change it this way: 3x + 2y = 5 3x + 2(something in terms of x) = 5 and then we will have one equation with one variable, which is the kind of thing we like to see, right? If only we had an equation like y = (something in terms of x) Where do you suppose we could get one? Well, if what we have to start with is two linear equations, then we can convert either one of them into such an equation! For example, 2x + y = 3 2x + y - 2x = 3 - 2x y = 3 - 2x Now we can substitute: 3x + 2y = 5 3x + 2(3 - 2x) = 5 \______/ y And now we're in business. Note that once you've solved for x, you can substitute that value back into y = (something in terms of x) to get the value of y. Note also that it doesn't matter which equation we convert: 3x + 2y = 5 2y = 5 - 3x y = (5 - 3x)/2 Also, it doesn't matter which variable we solve for: 3x + 2y = 5 3x = 5 - 2y x = (5 - 2y)/3 I picked the first equation to convert, and picked y as the variable to solve for, because y was already by itself in that equation, which meant that I wouldn't have to deal with fractions. As you solve more of these, you'll start to make up little tricks like that. But the point is, you can't really choose the 'wrong' equation or the 'wrong' variable. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/