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Solving Systems of Equations


Date: 10/30/96 at 11:52:48
From: sylvia hobson
Subject: algebra

I thought I knew how to do this, but guess I don't. 
Solve for a, b, and c in at least two ways:

2a+6b+c = 5
a+b-c = -1
a+8b+2c = 0
                                                                                                                                                              
Please help.                                                                                


Date: 10/31/96 at 17:40:57
From: Doctor Tracy
Subject: Re: algebra

Sylvia,

You are right that there are many approaches to solving this problem.  
They are really all equivalent mathematically, but everybody seems to 
prefer one approach or another.  I will show you how to solve your  
problem using two techniques and then you should be able solve other 
problems like this one.  

First of all, let's label your equations:

  (I)      2a + 6b + c = 5
 (II)       a +  b - c =-1
(III)       a + 8b +2c = 0

Technique 1:

The first thing that we want to do is eliminate one of the variables.  
Let's try to eliminate a.  Notice that if you add -2 times the second 
equation to the first equation, the a's will cancel:

(I)
          2a + 6b +  c =  5
 -2*(II) -2a - 2b + 2c =  2
 ----------------------------
           0 + 4b + 3c =  7

Do this kind of thing to eliminate a again.  I notice that equation 
(II) minus equation (III) will eliminate a and give me:

   0 - 7b - 3c = -1 ------> 7b + 3c = 1

Now we want to use our two new equations to eliminate either b or c.  
Since our first equation minus the second one will get rid of c, 
that's what I'll do. (If for some reason you really wanted to get rid 
of b at this point, you could multiply the first equation by 7 and the 
second equation by -4 and add.)  

Eliminating c we get -3b = 6, so b = -2.  Use this in one of the 
equations with only b and c to find c and then use your b and c in one 
of the original 3 equations to find a.

Technique 2:

First take one of the equations and solve for one of the variables:

  I will take equation (II) and solve for a to get: a = -1 - b + c.

Now, take one of the other equations and solve for the same variable:

  I will take equation (III) and solve for a to get: a = -8b - 2c.

Put these two together to get: -1 - b + c = -8b - 2c.

Now collect like terms to get: -1 + 7b + 3c = 0

Solve this equation for one variable: b = (1-3c)/7.

Replace b by this expression in one of the two equations for a above: 
i.e., take a = -1 - b + c and replace b to get:

  a = -1-[(1-3c)/7]+c -----> a = (6+10c)/7

Finally, go back to one of your original equations (I'll use (II)) and
replace a and b with the new expressions in c.  Take the equation 
a+b-c = -1 and replace a and b to get:

  (6+10c)/7 + (1-3c)/7 - c = -1.

Now, this is an equation in only c, so you should be able to solve it 
for c.  Once you have c, plug it in to the equations a = (6+10c)/7 and 
b = (1-3c)/7 to find a and b.

I hope this helps you.  

-Doctor Tracy,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra

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