A few days ago
challenge question! for math geniuses!?
what are all ordered triples of real numbers (x,y,z) that satisfy
(x+y)(x+y+z) =120
(y+z)(x+y+z)=96
(x+z)(x+y+z)=72
show the work.. if u get this question rite.. u r a math geniuss!
Sponsored question was the final question two competers had to face during the math olympiad of 06′. Winner solved it for a grand prize mitsu evo!
Top 6 Answers
A few days ago
Favorite Answer
First we can add these up to get:
2(x+y+z)^2 = 120 + 96 + 72 = 288
So x + y + z = +/-12
If x+y+z = 12, then x + y = 10 so z = 2
and similarly x = 4, and y = 6.
If x+y+z=-12 then z = -2, x = -4, y = -6
So the ordered triples that satisfy this are:
(4,6,2) and (-4,-6,-2)
1
4 years ago
in case you have a ti-eighty 4 you are able to resolve this with a matrix. purely circulate to the matrix menu, make a 5×6 matrix plugging interior the coefficients and solutions. Now circulate back to the homestead, circulate back to the matrix menu % rref( plug in [A] then press enter and you’re able to have your answer.
0
A few days ago
x=4
y=6
z=2
(4+6)(4+6+2)=120
(6+2)(4+6+2)=96
(4+2)(4+6+2)=72
0
A few days ago
ugh i thought i was close but noppe
it works for the first one
z=2
y=3
x=7
0
A few days ago
I’m going to go with (6,4,2)
This solves all equations.
If you want the work just ask.
0
A few days ago
Nope don’t have a clue!
0
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