Question

If $\rm \sqrt{a}x- \sqrt{b}y =b-a$ and $\rm \sqrt{b}x -\sqrt{a}y=0$ find the value of xy.​

Answer 1

Answer:

x=√a And y=√b

Step-by-step explanation:

hope it helps.

Answer 2

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm :\longmapsto\:\sqrt{a} \: x- \sqrt{b} \: y =b-a - - - - (1)$

and

$\rm :\longmapsto\:\sqrt{b} \: x -\sqrt{a} \: y=0 - - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:(\sqrt{a} + \sqrt{b}) x- (\sqrt{b} + \sqrt{a})y =b-a$

$\rm :\longmapsto\:( \sqrt{b} + \sqrt{a})(x - y) = {( \sqrt{b}) }^{2} - {( \sqrt{a})}^{2}$

$\rm :\longmapsto\:( \sqrt{b} + \sqrt{a})(x - y) = ( \sqrt{b} + \sqrt{a})( \sqrt{b} - \sqrt{a})$

$\bf\implies \:x - y = \sqrt{b} - \sqrt{a} - - - - (3)$

Now, On Subtracting equation (2) from (1), we get

$\rm :\longmapsto\:(\sqrt{a} - \sqrt{b}) x + (\sqrt{a} - \sqrt{b})y =b-a$

$\rm :\longmapsto\:(\sqrt{a} - \sqrt{b})(x + y) = {( \sqrt{b} )}^{2} - {( \sqrt{a} )}^{2}$

$\rm :\longmapsto\:(\sqrt{a} - \sqrt{b})(x + y) = ( \sqrt{b} + \sqrt{a})( \sqrt{b} - \sqrt{a})$

$\bf\implies \:x + y = - \sqrt{b} - \sqrt{a} - - - (4)$

Now, We know

$\rm :\longmapsto\:4xy = {(x + y)}^{2} - {(x - y)}^{2}$

On substituting the values from equation (3) and (4),

$\rm :\longmapsto\:4xy = {( \sqrt{b} + \sqrt{a} )}^{2} - {( \sqrt{b} - \sqrt{a} )}^{2}$

$\rm :\longmapsto\:4xy = 4 \sqrt{a} \sqrt{b}$

$\bf\implies \:xy = \sqrt{ab}$

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More Identities to know

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)