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Finding Solutions to Systems of Equations

Date: 10/18/96 at 22:34:58
From: Christian Sheehy
Subject: Finding Solutions to Systems of Equations

Dear Dr. Math,

I need help solving systems of equations!   My friends keep 
telling me what to do,  but I just can't get the hang of it :(


    Find the solution set of each equation:




Thanks for your help!   :)

Date: 10/19/96 at 11:5:54
From: Doctor Anthony
Subject: Re: Finding Solutions to Systems of Equations

Try to get the number in front of either m or n the same in both 

If we multiply the first equation throughout by 2, we shall 
get 10n in both equations.  When we subtract the equations, 
the term in n will disappear.

>     3m+5n=21  multiply along by 2  6m+10n = 42
>     4m+10n=38                      4m+10n = 38
                       subtract          2m = 4
                      divide by 2         m = 2

Return to one of the original equations and plug in m = 2    
     3 x 2 + 5n = 21    so  5n = 15    and  n = 3

So   m=2 and n=3.  Try these values in the second equation:
  4 x 2 + 10 x 3 = 38   
              38 = 38  Yep!  So these values are correct.

>     -2x+9y=5       multiply by 4    -8x+36y = 20 
>     8x-36y=-20                       8x-36y = -20

If we add these equations both x and y terms go out.  If we look 
closely, we can see that in fact the two equations are really the 
same equation. If you multiply the second equation by -1 throughout, 
you get the first equation.  In this case there is no single point 
(x,y) satisfying the equations. We have a 'solution set' which is 
all points on the line -2x+9y=5. So there are an infinite number of 

>     x+5y-2z = 5
>     2x+y-z  = 4

These are the equations of two planes, and the solution set is the 
line of intersection of the planes.  If you are familiar with 3 
dimensional work, we first write the equations in terms of any common 
point that we like to choose. For example the point (1,0,-2) lies on 
both planes, so we could write the equations:

     (x-1) + 5(y) - 2(z+2) = 0
    2(x-1) +   y  -  (z+2) = 0

Now using determinants the equation of the line can be written

     (x-1)     -y      (z+2)
     -----  = ----- =  -----
    |5  -2|  |1  -2|  |1   5|
    |1  -1|  |2  -1|  |2   1|

     (x-1)     -y      (z+2)
     -----  = -----  = ------
       -3       3        -9

     (x-1)     y       (z+2)
     -----  = -----  = ------  = k
       1       1         3 

So    x = 1 + k
      y = k
      z = -2 + 3k

We can check for any other point on the common line of the two planes.  
Try k = 1. This gives x=2, y=1, z=1. If you put these values in the 
equations of the planes you have 
  x+5y-2z = 5     2+5-2 = 5   Yes OK
 2x+y- z = 4      4+1-1 = 4   Yes OK

So the expressions for x, y, z in terms of k (any value you like) will 
give the solution set for these equations.  Again there are an 
infinite number of points that lie on the line of intersection of the 

This last question was a lot more advanced than the earlier ones, so I 
expect they just wanted you to realize that two equations with three 
unknowns, (x,y,z) will not give you a single solution.

-Doctor Anthony,  The Math Forum
 Check out our web site!   

Date: 10/18/2002 at 13:48:19
From: Judy Foster
Subject: Solutions to Systems of Equations

Dear Dr. Math:

In your answer for

   3m + 5n = 21
  4m + 10n = 38

you suggested using 2 in the first equation. Should it have been -2 
in order to cancel out 10n?  I'm in college and learning myself and 
wanted to be sure I was thinking correctly.  

Thank you.

Date: 10/18/2002 at 23:28:04
From: Doctor Peterson
Subject: Re: Solutions to Systems of Equations

Hi, Judy.

In this example, you are told to multiply the first equation by 2, 
and then to SUBTRACT the equations. That is the same as multiplying 
by -2 and ADDING, which is what you want to do. So your thinking is 
fine. In fact, when I work these out on paper I usually indicate each 
step I do by showing the numbers I multiply by before adding; this 
can be taken instead as the positive multiplier together with a sign 
indicating whether to add or subtract:

    3m +  5n = 21
                  \ 2
                     2m = 4  m = 2
    4m + 10n = 38

Here I chose to multiply the first by 2 and the second by -1 (that 
is, to subtract the second) so I would get a positive coefficient of 
m; but you could just as well have used -2 and 1 to get -2m=-4. I 
drew a line from each equation to the place where I planned to put 
the new combined equation, and wrote the multipliers next to them to 
remind me what to multiply by; then I multiplied and subtracted in 
one step to save writing.

But that trick is just a little extra for you. The answer to your 
question is, you're doing fine!

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
Associated Topics:
High School Basic Algebra
High School Linear Algebra

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