Question

Find the value of x and y by the method of cross-multiplication:

$\frac{ {a}^{2} }{x} - \frac{ {b}^{2} }{y} = 0 \\ \\ \frac{ {a}^{2} {b}^{2} } {x} + \frac{ {b}^{2} a}{y} = a + b$
Given a condition that :- x and y are not equal to 0

$\textbf{Class : X}$

Answer only if uh know ! No spam answers needed here !​

Answer 1

So, In this question, we have to use the Cross multiplication method.

Let's assume 1/x as u and 1/y as v.

Hence, equations obtained :

$\tt{ {a}^{2} u - {b}^{2} v = 0}$

Here,

we have :

a1 = a^2

b1 = b^2

c1 = 0

and

$\tt{a {}^{2} bu + {b}^{2} av - (a + b) = 0}$

a2 = a^2

b2 = b^2a

c2 = - ( a + b )

By cross multiplying, we get :

$\tt{ = \frac{ {u}^{2} }{b {}^{2}(a + b) - 0 \times {b}^{2}a} }$

${\tt{ \frac{ - v}{ - {a}^{2}(a + b) - 0 \times {a}^{2} b} } = \frac{1}{ {a}^{3} {b}^{2} + {a}^{2} b {}^{3}} }$

$\tt{ \frac{ {u}^{2} }{b {}^{2} (a + b)} = \frac{v {}^{2} }{ {a}^{2} (a + b)} = \frac{1}{ {a}^{2} {b}^{2}(a + b)} }$

Now,

$\tt{ \frac{u}{b {}^{2} (a + b)} = \frac{1}{ {a}^{2} {b}^{2}(a + b)} }$

Hence,

$\tt{u = \frac{b {}^{2} (a + b)}{ {a}^{2}b {}^{2} (a + b)} }$

Thus,

u = 1/a^2 ... (1)

We got value of u as 1/a^2,

v = ?

Now,

$\tt{\frac{v}{ {a}^{2} (a + b)} = \frac{1}{a {}^{2} {b}^{2} (a + b)}}$

Hence,

$\tt{v = \frac{ {a}^{2} (a + b)}{ {a}^{2} b{ }^{2} (a + b)}}$

Thus,

v = 1/b^2 ... (2)

From (1) and (2),

x = 1/u = a^2

y = 1/v = b^2

Thus,

x = a²

y = b²

________________________

Hope it helps!

Answer 2

Correction in question :-

$\frac{ {a}^{2} }{x} - \frac{ {b}^{2} }{y} = 0 \\ \\ \frac{ {a}^{2}b} {x} + \frac{ {b}^{2} a}{y} = a + b$


So,


$\boxed{ \boxed{ x = \frac{b1c2 - b2c1}{a1b2 - a2b1}}}$

$\boxed{ \boxed{y = \frac{c1a2 - c2a1}{a1b2 - a2b1}}}$

Now,

In the question :-

Let :-

$\frac{1}{x} \: be \: u \\ \\ \\ \frac{1}{y} \: be \: v$

So,

We have the two equations as follows :-

$\boxed{ \boxed{ \sf{ {a}^{2}u - {b}^{2}v + 0 = 0}}}$

$\boxed{ \boxed{ \sf{ {a}^{2} bu + a{b}^{2}v - (a + b) = 0}}}$

So,

Here, considering two equations and putting in the formula of cross multiplication method mentioned above :-

So,

$\boxed{u = \frac{ {( - b)}^{2}( - a - b) - a {b}^{2} \times 0}{ {a}^{2}(a {b}^{2}) - {a}^{2}b{( - b)}^{2}}} \\ \\ \\ \boxed{u = \frac{a {b}^{2} + {b}^{3}}{ {a}^{3} {b}^{2} + {a}^{2} {b}^{3}}}$

On simplifying we get

$x = \dfrac{ {a}^{3} + {a}^{2}b}{a + b}$

$\boxed{x = {a}^{2}}$

Similarly we get

$y ={b}^{2}$