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Substitution and Elimination


Date: 01/13/2002 at 15:49:00
From: Kathryn
Subject: Substitution and Elimination

I don't understand substitution and elimination. Could you please 
explain them to me step by step? 

Thanks!


Date: 01/13/2002 at 19:47:20
From: Doctor Ian
Subject: Re: Substitution and Elimination

Hi Kathryn,

Suppose I have these two equations:

  3x + y = 9

  4x - y = 5

Elimination works by adding or subtracting equations to get rid of one 
of the variables. In this case, I can add the two equations:

  (3x + y) + (4x - y) = 9 + 5

Do you see why this works? This is really just an application of the 
basic rule in algebra, which is that I can add the same thing to both 
sides of any equation. That is, 

  3x + y + [something] = 9 + [the same thing]

In this case, I'm just writing the thing in two different ways. Make 
sure that you see why this is legal. If you don't, it can seem like 
magic. 

Anyway, after adding the equations, I notice that y conveniently 
disappears:

  3x + y + 4x - y = 9 + 5

  3x + 4x + y - y = 14

               7x = 14

                x = 2

Now that I know the value of x, I can plug it back into either 
equation to get the corresponding value of y.  

Now, what if things didn't work out so conveniently? What if I'd had 
these equations instead?

  3x + 2y = 12

  4x -  y =  5

Well, in this case, I can still use elimination, but I'd have to scale 
one of the the equations to get back to a situation where addition or 
subtraction would work. In this case, I can multiply everything in the 
second equation by 2 (do you see why this works?) to get

  3x + 2y = 12

  8x - 2y = 10

and now I'm more or less back where I was before. Sometimes it may be 
necessary to scale both equations, e.g., 

  3x + 2y = ...    

  5x - 3y = ...

Here, you'd want to multiply the first equation by 3, and the second 
by 2, to get rid of the y terms; or you could multiply the first 
equation by 5, and the second by 3, to get rid of the x terms. It's 
very similar to what you have to go through to find a common 
denominator in order to add fractions.  


And sometimes you may have to subtract one equation from the other, 
rather than adding them. But these are just little tricks that you 
have to use. The main thing is that you need to have an identical term 
(e.g., 2y) in both equations, so that you can get the variable in that 
term to disappear. 

What about substitution? Well, let's look at these equations again:

  3x + 2y = 12

  4x -  y =  5

Note that I can rewrite either equation to look like y = ...:

  3x + 2y = 12                       4x - y = 5
 
       2y = 12 - 3x                  4x - 5 = y

        y = 6 - (3/2)x         

Or I could rewrite them to look like x = ... It doesn't really matter.  
The important thing is that I now have an expression for y in terms of 
x (or an expression for x in terms of y). So wherever I see the 
variable, I can substitute the expression... for example, in the other 
equation.

        3x + 2y = 12                 but y = 4x - 5

 3x + 2(4x - 5) = 12                 so substitute for y

   3x + 8x - 10 = 12                 and solve for x

            11x = 22

              x = 2
  
As before, I can now use the value of x to get the value of y by 
plugging it into either equation. 

This will all make more sense if you have a solid understanding of 
what it means to solve a system of linear equations. To find a simple 
explanation of that, see

   The Idea behind Simultaneous Equations
   http://mathforum.org/library/drmath/view/53285.html   

Does this help?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/   


Date: 01/14/2002 at 15:22:27
From: Kathryn
Subject: Substitution and Elimination

Thanks soooo much! I really understand it now! I am so happy!! 
Thanks so much for explaining it to me!

Kathryn
    
Associated Topics:
Middle School Algebra
Middle School Equations

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