Question

If x+1/y=4 and y+1/z=1 and z+1/x=7/3 find xyz

Answer 1

Given,
x + 1/y = 4 --------(1)
y + 1/z = 1 ---------(2)
z + 1/x = 7/3 -------(3)

from equation (1), x + 1/y = 4
⇒ 1/y = 4 - x
put it in equation (2),
1/(4 - x) + 1/z = 1
⇒1/(4 - x) - 1 = - 1/z
⇒(1 - 4 + x)/(4 - x) = -1/z
⇒ (3 - x)/(4 - x) = 1/z
⇒ z = (4 - x)/(3 - x) , put it in equation (3)

(4 - x)/(3 - x) + 1/x = 7/3
⇒{(4 - x)x + (3 - x)}/x(3 - x) = 7/3
⇒ {4x - x² + 3 - x }/(3x - x²) = 7/3
⇒(-x² + 3x + 3 )/(3x - x²) = 7/3
⇒-3x² + 9x + 9 = 21x - 7x²
⇒4x² - 12x + 9 = 0
⇒ (2x - 3)² = 0 ⇒ x = 3/2

put x = 3/2 in equation (3),
z + 2/3 = 7/3
z = 5/3
Now, z = 5/3 put in equation (2),
y + 3/5 = 1
y = 2/5

Hence, xyz = 3/2 × 5/3 × 2/5 = 1
xyz = 1

Answer 2

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