Please solve this problem for me !?

Question

x , y , z = ?
( 7 / x + y )=( 6 / x + z )=( 9 / y + z )
and ( x + y )( Y + z )( z + x ) = 24192

( x + y )=( y + z )=( z + x )
sorry for the Y . It was y

sorry it is
( x + y )( y + z )( z + x )=24192

Answer 1

I think is:

7 / (x + y )= 6 / (x + z )= 9 /( y + z )

7 / (x + y )= 6 /( x + z )
6*(x+y) = 7*(x+z)
(x+z) = 6/7*(x+y) (1)

6 / (x + z )= 9 /( y + z )
9*(x+z) = 6*(y+z)
(x+z) = (6/9)*(y+z)
(x+z) = (2/3)*(y+z)(2)

From (1) and (2):
(x+z) = 6/7*(x+y) = (2/3)*(y+z)
6/7*(x+y) = (2/3)*(y+z)
(x+y) = ((2/3)*(y+z))/ (6/7)
(x+y) = ((2/3)*(y+z))*(7/6)
(x+y) = (14/18)*(y+z)
(x+y) = (7/9)*(y+z) (3)


( x + y )( y + z )( z + x ) = 24192

From (2) and (3):

((7/9)*(y+z))* (y+z)* ((2/3)*(y+z)) = 24192
(7(y+z)*(y+z)*2(y+z))/(3*9) = 24192
(7(y+z)*(y+z)*2(y+z))/27 = 24192
(7(y+z)*(y+z)*2(y+z)) = 24192*27
(14*(y+z)³) = 24192*27
(y+z)³ = (24192*27)/14
(y+z)³ = (24192*27)/14
(y+z)³ = 1728*27
(y+z)³ = 1728*3³
(y+z)³ = (2^6*3³)*3³
(y+z)³ = 2^6*3^(3+3)
(y+z)³ = 2^6*3^6
(y+z) = 2²*3²
(y+z) = 4*9
(y + z) = 36 (4)

(4) in (2):

(x+z) = (2/3)*(y+z)
(x+z) = (2/3)*36
(x+z) = 2*12
(x+z) = 24 (5)

(4) in (3):

(x+y) = (7/9)*(y+z)
(x+y) = (7/9)*36
(x+y) = 7*4
(x+y) = 28 (6)


From (4), (5), (6):
(y+z) = 36 (4)
(x+z) = 24 (5)
(x+y) = 28 (6)

(4) - (5)

y - x = 36 - 24
y - x = 12
y = 12 + x (7)

(7) in (6):
(x+y) = 28
(x+12 + x ) = 28
2x+12 = 28
2x = 16
x = 8 (8)

(8) in (7):
y = 12 + x
y = 12+8
y = 20 (9)

(9) in (4):

(y+z) = 36
20 + z = 36
z = 16 (10)

From (8), (9) and (10):
x = 8 (8)
y = 20 (9)
z = 16 (10)


Conferring :

7 / (x + y )= 6 / (x + z )= 9 /( y + z )
7 / 28 = 6 / 24 = 9 /36
1/4 = 1/4 = 1/4 (OK)

( x + y )( y + z )( z + x ) =
28*36*24 =
1008*24 =
24192 (OK)

Answer: x = 8; y = 20 ; z = 16


Kisses from Brazil

=**