Question

10. If x^2 + y^2 = 45 and xy = 18, then find the value of 1/x+1/y
(a) 1/2
(b)1/3
(c) 1/4
(d) 2/3

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Answer 1

Good Morning!!

x² + y² = 45 ( Given )

xy = 18 ( Given )

1/x + 1/y = ?

x² + y² = 45

( x + y )² - 2xy = 45

( x + y )² = 45 + 2 (18 )

( x + y ) = ±√(81)

x + y = ±9

1/x + 1/y = ?

? = ( x + y )/xy

? = ±9/18

? = ±1/2

So, 1/x + 1/y = ±1/2

Answer 2

The value is $\frac{1}{4}$, hence the correct option is (c).

Step-by-step explanation:

Given:

x²+y²=45

xy=18

$\frac{1}{x}+\frac{1}{y} \\\\=\frac{y+x}{xy}$

Taking square

$=\frac{(y+x)^{2} }{(xy)^{2} } \\\\=\frac{x^{2}+y^{2}+2xy }{(xy)^{2} }$

Substituting the known values

$=\frac{45+(2\times 18)}{18\times 18} \\\\=\frac{45+36}{324} \\\\=\frac{81}{324} \\\\=\frac{1}{4}$

Learn More:

If X + Y + Z is equal to 1, xy + yz plus zx is equal to -1 and x y z is equal to -1, find the value of x cube + y cube + Z cube

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