Suppose that x , y , z are three positive numbers that satisfy the equation x y z

Quintin Stafford 2022-06-26 Answered
Suppose that x , y , z are three positive numbers that satisfy the equation x y z = 1 , x + 1 z = 5 and y + 1 x = 29. Then z + 1 y = m n , where m and n are coprime. Find m + n + 1
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Answers (2)

grcalia1
Answered 2022-06-27 Author has 23 answers
Just substitute the values of y and z in x y z = 1
x ( 29 x 1 ) x ( 5 x ) = 1
29 x 1 = 5 x
x = 1 5
y = 29 / 5 1 1 / 5 = 24
z = 1 5 1 / 5 = 5 24
z + 1 y = 5 24 + 1 24 = 1 4
Thus,
m + n + 1 = 1 + 4 + 1 = 6

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Garrett Black
Answered 2022-06-28 Author has 5 answers
Another solution: we have 1 x = y z, 1 y = x z, and 1 z = x y. So
x + x y = 5 y + y z = 29 z + x z = Q
where Q is the quantity we're trying to determine.
Multiplying these three equations together gives 145 Q = x y z ( x + 1 ) ( y + 1 ) ( z + 1 ) and hence 145 Q = ( x + 1 ) ( y + 1 ) ( z + 1 )
Expanding, we have
( x + 1 ) ( y + 1 ) ( z + 1 ) = x y z + x y + y z + x z + x + y + z + 1 = 2 + x y + y z + x z + x + y + z = 2 + 5 + 29 + Q = 36 + Q
and so 145 Q = 36 + Q, from which it follows that Q = 1 4

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