+876 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
https://web2.0calc.com/questions/some-of-my-questions-got-deleted-question
hope this helped Jimmy neutron
+2407 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.
=^._.^=
+876 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
