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3 Equations, 3 Variables


Date: 01/26/98 at 15:20:34
From: Greg & Kelly Remor
Subject: Solving 3 Equations in 3 Variables

Dr. Math:

I am really rusty with my algebra problems.  How do you solve a system 
of three equations in three variables?

 1121 = 25a + 5b + c
  626 = 49a + 7b + c
  967 = 81a + 9b + c 

The answers are: a = 104.5   b = -1501.5   c = 6016

Thanks for your help!


Date: 01/28/98 at 12:54:18
From: Doctor Jaffee
Subject: Re: Solving 3 Equations in 3 Variables

Dear Greg and Kelly,

This problem looks much like the problems in which you are given 
three points on a parabola and you have to calculate the equation 
of the parabola. In this case the points would be (5,1121),(7,626), 
and (9,967). Since the general equation of a parabola is 
y = ax^2 + bx + c, if you substitute the x and the y numbers into 
this equation for each pair of numbers, the result will be exactly 
the three equations you submitted. But maybe this is more information 
than you really want to know. Let's get down to actually solving the 
system.

The easiest method I know is the "elimination method". What we want to 
do is combine two of the equations in such a way that one of the 
variables is eliminated, then take another pair of the original 
equations and combine them in a way to eliminate the same variable.  
I'll show you what I mean.

Since there is a "c" at the end of each equation, that would be the 
easiest variable to eliminate.

So, take the first two equations,  1121 = 25a + 5b + c
                                   626 = 49a + 7b + c

When you multiply the second equation through by -1 you get

                                   -626 = -49a + -7b + -c 

and when you add that to the first equation you get

                                   *495 = -24a + -2b

Then work with the 2nd and 3rd equations:

                                     626 = 49a + 7b + c
                                     967 = 81a + 9b + c

multiply the first equation through by -1 and get      

                                    -626 = -49a + -7b + -c

then add that to the second equation with the result 

                                    *341 =  32a + 2b

Now the two equations with the asterisks only have two variables and 
we can eliminate the "b" by adding these two equations:

                                     495 = -24a + -2b
                                     341 =  32a +  2b 
                                     836 =   8a

Now divide both sides by 8 and the result is a = 104.5. Substitute 
that back into one of the equations that just has the variables 
a and b and you should get b = -1501.5.

Finally, go back to the original equations, pick any one, and 
substitute your values for a and b and you will get c = 6016.

Now getting back to the problem that I discussed originally, the 
equation of the parabola would be y = 104.5x^2 + -1501.5x + 6016.

I hope I've helped in clearing off some of your rust.  Good Luck.   

-Doctor Jaffee,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
Middle School Equations

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