+0  
 
0
148
2
avatar

i dont know how to solve this 3 variable system of equations and i keep getting stuck 
x+y+z=8
x+4y+6z=24 

 Sep 15, 2019
 #1
avatar+106536 
+1

We have more variables than equations....so.....we will have  infinite solutions

 

Subtracting the first equation from the second  we get that  3y + 5z  =  16  → 

 

 3y = [ 16 - 5z]  →     y =  [ 16 - 5z ] / 3 

 

And using the first equation

 

x + y + z =  8

x + [ 16 - 5z]/3 + z  = 8       multiply through by 3

 

3x + 16 -5z + 3z = 24

3x - 2z = 8

3x = 8 - 2z

x = [ 8-2z] /3

 

So....the solutions for  {x, y , z }  =   {  [8-2z]/3 , [16-5z]/3 , z }

 

For example...if z = 0,  then x = 8/3  and y = 16/3

 

So....check that these solutions work :

 

x + y + z  = 8  ???

8/3 + 16/3 + 0  = 8 ???

24/3 = 8  ???

True

 

And

 

x + 4y + 6z = 24  ???

(8/3) + 4 [16/3]  + 0  = 24??

8/3 + 64/3  = 24 ???

72/3  = 24   ???

True

 

So.....   one possible solution is   { x, y , z}  =  (8/3, 16/3 , 0 }

 

 

cool cool cool

 Sep 15, 2019
edited by CPhill  Sep 15, 2019
 #2
avatar+507 
+2

x=2n, y=12-5n, z=3n-4 for any n

 Sep 15, 2019

11 Online Users