+0  
 
0
61
1
avatar

In this multi-part problem, we will consider this system of simultaneous equations:  

               3x+5y-6z=2, (i)  

              5xy-10yz-6xz=-41,  (ii)    

              xyz=6.  (iii)    

Let a=3x, b=5y, and c=-6z.  

 

i figured out a so i won't post it

 

(b) Given that (x,y,z) is a solution to the original system of equations, determine all distinct possible values of x+y.

i'm only finding a few values of x+y, how do i find all of them?

 Jun 6, 2020
 #1
avatar+807 
+4

Replace x, y, and z in equations by a, b, and c:

a + b + c = 2

 

a*b/3 + b*c/3 + a*c/3 = -41.  So.  a*b + b*c + a*c = -123

 

-a*b*c/(3*5*6) = 6. So. a*b*c = -540

 

(t - a)(t - b)(t - c) = 0. ⇒ t^3 - (a + b + c)t^2 + (a*b + b*c + a*c)t - a*b*c = 0

 

Hence:  t^3 - 2t^2 - 123t + 540 = 0

 

Solving this gives: a = -12, b = 5, c = 9 or variants of these.

 

So:  x = a/3 → -4,   y = b/5 → 1,  z = -c/6 → -3/2

 

x + y = -3 This is just one possibility.  The others can be found by interchanging the values allocated to a, b and c.

Hope this helps, whymenotsmart:)

 Jun 6, 2020

19 Online Users

avatar
avatar