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0
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avatar+742 

Greetings,

The problem goes like this:

Find all values of x+y, given

3x+5y-6z=2

5xy-10yz-6xz=-41

xyz=6

 

What I did is I let a=3x, b=5y, and c=-6z.

I let a,b,c be the roots of a poly, and got the poly to be n^3 - 2n^2 - 123n + 540

 

Sicne it asked for x+y, I let it be in terms of a and b. i found that x+y = (5a+3b)/15.

I used the rational root theorem and used synthetic division one by one of the potential roots, and The poly factored into (t-5)(t-9)(t+12)

 

So we have the potential

a=5,b=9;

a=5, b= -12;

a=9, b=5;

a=9, b=-12;

a=-12, b=5;

a=-12, b=9;

 

And for these, I got 

52/15, -11/15, 4, 3/5, -3, 11/5 when plugging in to my equation

 

but it said that I was wrong. I was wondering why. Then i put it into wolfram and it gave  52/15, -11/15, 3/5, -3, 11/5

 

this was also wrong as I expected, since computer may miss a few pointers.

 

Can anyone show me where I Was wrong? Thanks in advance!

 

Best,

Jimmy

 May 29, 2021
 #1
avatar
0

https://web2.0calc.com/questions/some-of-my-questions-got-deleted-question

 

hope this helped Jimmy neutron

 May 29, 2021
 #2
avatar+2127 
+1

Setting a, b, and c is really smart. 

a = 3x

b = 5y

c = -6z

 

3x + 5y - 6z = 2

a + b + c = 2

 

5xy - 10yz - 6xz = -41

ab + bc + ac = -123

 

xyz = 6

abc = -540

 

x^3 - 2x^2 - 123x + 540

(x - 5)(x - 9)(x + 12)  # I don't know how to factor cubic equations, I just plugged it into a calculator. 

x = 5, 9, -12

 

a = 3x

b = 5y

c = -6z

 

a = 5, b = 9

5/3 + 9/5 = 52/15

a = 5, b = -12

5/3 - 12/5 = -11/15

a = 9, b = 5

9/3 + 5/5 = 4

a = 9, b = -12

9/3 - 12/5 = 3/5

a = -12, b = 5

-12/3 + 5/5 = -3

a = -12, b = 9

-12/3 + 9/5 = -11/5

 

Everything looks right to me, I'm not sure why it's wrong. :((

Looking at Asinus's answer from the link guest sent, he had the same aproach too. 

 

=^._.^=

 May 29, 2021
 #4
avatar+742 
+1

Thanks for your post and thoughtful time! 

 

And how I factor a cubic equation, is I use the rational root theorem to find possible roots. Then I use synthetic division one by one to rule out the false ones. Then it just makes it into (x-n)(x-m)... where n,m,etc. are my roots

MathProblemSolver101  May 30, 2021
 #5
avatar+2127 
+1

Ohh, I never knew about the rational root theorem. 

http://wmueller.com/precalculus/families/ratroot.html

That's quite smart. :))

 

=^._.^=

catmg  Jun 1, 2021
 #3
avatar+121004 
+1

Here's  what  WolramAlpha  shows   :

 

https://www.wolframalpha.com/input/?i=3x%2B5y-6z%3D2%2C++5xy-10yz-6xz%3D-41%2C++xyz%3D6

 

Apparently,  more  than one possible  value  for   x  +  y  ......

 

 

cool cool cool

 May 29, 2021

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