Solve 2 by root x + 3 by root y = 2 ; 4 by root x – 9 by root y = -1, x, y o

Question

Q) Solve :-   $\frac{2}{\sqrt X} + \frac{3}{\sqrt Y}$ = 2; $\frac{4}{\sqrt X} - \frac{9}{\sqrt Y}$ = - 1, x, y > o

Sapling Academy · Accepted Answer

Let's consider, $\frac{1}{\sqrt X}$  = A and $\frac{1}{\sqrt Y}$ = B

therefore the given equations will be:

Equation $\frac{2}{\sqrt X} + \frac{3}{\sqrt Y}$ = 2 will become:

2 A + 3 B = 2 .......... (i)

Similarly, $\frac{4}{\sqrt X} - \frac{9}{\sqrt Y}$ = - 1 will become:

4 A - 9 B = -1 ..... (ii)

By solving equations (i) and (ii), we get:

A = 1/2 and B = 1 /3

Next, let's put these values back into original form of A and B, Hence,

A = $\frac{1}{\sqrt X} = \frac{1}{2}$

By squaring on both sides, we get:

$\frac{1}{X} = \frac{1}{4}$

X = 4

Similarly, B = $\frac{1}{\sqrt Y} = \frac{1}{3}$

By squaring on both sides, we get:

$\frac{1}{Y} = \frac{1}{9}$

Y = 9

Since both values satisfies the given condition of X, Y > 0.

Therefore value of X is 4 and value of Y is 9.

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