Help me please (´□｀。)

Question

Solve the following simultaneous equations but x≤y≤z

x+y+z = 6
x²+y²+z² = 38
x³+y³+z³ = 144

what should I do？

Answer 1

 

 I would solve it numerically or equivalently examine their graphs.

 

If an explicit algebraic solution is possible, it will take much

 longer to find it than to do it numerically.

 

You will probably have to create a throw away program

to do the number crunching for you. Here's some psuedo code...

 

 Using the programming language of your choice, write three functions

  F1 ( x, y) ,  f1(x,y) and F2(x,y)   where x and y are double valued floating point vars

  

 The first does the following:

             z = 6 - x - y

            return ( x*x + y*y + z*z)

 

The second does the following:  <--- negative square root case

    return ( - F1(x,y))

 

 

The third does the following

          z = 6 - x - y

          return ( x*x*x+y*y*y+z*z*z)

 

call these two functions inside of a nested loop, which supervises

and monitors their return values. You are looking for the

results of the first function to be 38 and the second to be 144.

You can set a threshold, that these values be within a specific

tolerance or degree of accuracy like 0.25 or 1/2.

 

Something like this:

 

  x = -100

  y = -100

  Hx = 0.1   <--- increments x and y by a value of your choice

  Hy = 0.1   <--- the smaller you make this value the more accurate it will be

                  <--- but will take longer 

 

   while x <=100

       while y <= 100

       

             result1 = F1(x,y)

             Result1 = f1(x,y)

             result2 = F2(x,y)

 

             if (37.5 < result1 < 38.5) and (143.5 < result2 < 144.5)  OR

                (37.5 < Result1 < 38.5) and (143.5 < result2 < 144.5) 

             then output(x,y)

             

             y = y+Hy

         end while 

 

           x = x + Hx

   end while

----------------------------------------------------

 

  Your other option is to find a 2-D function plotter online and

  examine the graphs of   Y1 = f(x,y) = sqrt(38 - x^y - y^2)

 

           and Y2 = f(x,y) = cube root ( 144- x^3 - y^3)

 

                and Y3 = 6 - x - y

Answer 2

The only solution (positive integers) for the middle equation is (2, 3, 5). 

This does not satisfy the top equation. Forget about the bottom one.