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Solving Equations

Date: 01/15/98 at 09:12:34
From: Anonymous
Subject: Unsolved problem

Dear Dr. Math,

Please help me solve these equations:

   x2+y = 31
   y2+x = 41

The solution must be pure algebra, without using a graphical 
calculator. If you can, please show me the whole process.

The Farkash Family

Date: 01/15/98 at 14:05:04
From: Doctor Wilkinson
Subject: Re: Unsolved problem

When I look at these equations, I see that if I do the obvious thing 
and substitute y = 31 - x^2 from the first equation into the second 
equation, I'm going to get an equation with x appearing to the fourth 
power, which doesn't look good to me, so I start looking for a 

Maybe there's a solution in whole numbers and maybe I can just guess 
the answer!

There aren't very many numbers whose square is less that 31. Suppose I 
try x = 5. Then x squared is 25 and y is 6. And it works, because then 
y squared is 36, and y squared + x is 41.

I lucked out a little by guessing right on my first try, so let's 
suppose I hadn't been so lucky.

If I start by trying x = 0, I get y = 31, and y squared would be much 
too large. In fact, y must be less than 7 and x must be less than 6.  
If I try 

    x = 1, I get y = 30
    x = 2, I get y = 27
    x = 3, I get y = 22
    x = 4, I get y = 15

So x = 5 is the only possibility, and as we've seen, it works.

Sometimes trial and error is a good method if you can narrow down the
choices enough.
-Doctor Wilkinson,  The Math Forum
 Check out our web site!   

Date: 01/15/98 at 14:11:14
From: Doctor Rob
Subject: Re: Unsolved problem

Solve the first for y, and substitute the result into the second:

                        y = 31 - x^2,
           (31-x^2)^2 + x = 41,
   x^4 - 62*x^2 + x + 920 = 0.

Now we have to factor this 4-th degree polynomial in x:

   (x - 5)*(x^3 + 5*x^2 - 37*x - 184) = 0.

Clearly one solution is x = 5, so y = 6.  

There are three other solutions which come from the three roots of 
x^3 + 5*x^2 - 37*x - 184 = 0.  They are all real numbers, and are 
about 6.075336, -4.921732, and -6.153604.  The corresponding values of 
y are about -6.866848, 6.776558, and -5.909709. Their exact expression 
can be found as follows.

To get rid of the quadratic term, let w = x + 5/3, so that

   w^3 - (136/3)*w - 3053/27 = 0.

Now look for solutions of the form w = z + 136/(9*z).  Then

   z^3 - 3053/27 + 2515456/(729*z^3) = 0,
   z^6 - (3053/27)*z^3 + 2515456/729 = 0.

This is a quadratic equation in z^3 which you can solve for z^3:

   z^3 = (3053/27 + Sqrt[-27445/27])/2.

This has three roots, z1, z2, and z3.

   z1 = Cuberoot[(3053/27 + Sqrt[-27445/27])/2],
   z2 = z1*(-1+Sqrt[-3])/2,
   z3 = z1*(-1-Sqrt[-3])/2.

Each of them give a value of w,

   w1 = z1 + 136/(9*z1),
   w2 = z2 + 136/(9*z2),
   w3 = z3 + 136/(9*z3),

and then a value of x,

   x1 = -5/3 + z1 + 136/(9*z1),
   x2 = -5/3 + z2 + 136/(9*z2),
   x3 = -5/3 + z3 + 136/(9*z3).

The miracle is that all of the complex numbers drop out of the 
expressions for w1, w2, w3, x1, x2, and x3, so that they are real 

Possibly the question required positive values of both x and y.  
Another possibility is that the question required either integer or 
rational values of x and y.  In either case, there is just the one 
solution, x = 5, y = 6.

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

Date: 01/16/98 at 06:36:16
From: Anonymous
Subject: Thank You Very Much!

Dear Dr. Rob and Dr. Wilkinson,

Thank you very much for helping me solve my equation.
I think your site is a great idea and I am sure I will use your 
help again in the future.

Sincerely yours,
Mickey Farkash
Associated Topics:
Middle School Algebra

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