if 37/13=2+1/x+1/y+1/z find x, y and z
Answer (1)
Student Expert
11th Sep, 2019
Hello there!
Greetings!
We have,
37/13 = 2+1/(x+1/(y+1/z))
2 + 11/13 = 2 + 1/(x+1/(y+1/z))
===>11/13 = 1/(x+1/(y+1/z))
Taking reciprocal on both sides
13/11 = x+1/(y+1/z)
1 + 2/11 = x + 1/(y+1/z)
===> x=1 and 1/(y+1/z) = 2/11
1/(y+1/z) = 2/11
Taking reciprocal
y+1/z = 11/2
y + 1/z = 5 + 1/2
===> y = 5
1/z = 1/2
z = 2.
Thus,x = 1, y=5, z=2 .
Thankyou
Greetings!
We have,
37/13 = 2+1/(x+1/(y+1/z))
2 + 11/13 = 2 + 1/(x+1/(y+1/z))
===>11/13 = 1/(x+1/(y+1/z))
Taking reciprocal on both sides
13/11 = x+1/(y+1/z)
1 + 2/11 = x + 1/(y+1/z)
===> x=1 and 1/(y+1/z) = 2/11
1/(y+1/z) = 2/11
Taking reciprocal
y+1/z = 11/2
y + 1/z = 5 + 1/2
===> y = 5
1/z = 1/2
z = 2.
Thus,x = 1, y=5, z=2 .
Thankyou
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