The set of integers {x,y,z} complies with the following set of equations:
x+y+xy=17
x+z+xz=35
y+z+yz=71
Evaluate {x,y,z}.
Simon's Favorite Factoring Trick
(x+1)(y+1) = 18
(x+1)(z+1) = 36
(y+1)(y+1) = 72
let a=x+1, b=y+1, c=z+1
1) ab = 18 = 3*3*2
2) ac = 36 = 3*3*2*2
3) bc = 72 = 3*3*2*2*2
Since multiplying any two of these yields a positive number,
a,b,c must either all be positive or all negative
From equations 1 and 2: c = 2b and
from 2 and 3: b = 2a; so c = 4a
a*2a = 2a^2 = 18, a = ±3
b = 2a = ±6
c = 4a = ±12
(a,b,c) = {(3,6,12) or (-3,-6,-12)}
Subtract one from each:
So (x,y,z) = {(2,5,11), (-4,-7,-13)}
{edited to correct an arithmetic error}
Edited on September 19, 2023, 10:00 am
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Posted by Larry
on 2023-09-18 14:17:13 |
solution
