Question

if bx + ay = a2 + b2 & ax- by = 0 then the value of x-y is........

Answer 1

(b-a) (a2+b2)/(a sq+b sq) is the answer

Answer 2

Step-by-step explanation:

$b \times \: + \: ay \: = a {}^{2} + b {}^{2}.......(1)$

$ax \: - \: by \: = 0......(2)$

$Now \: \\ re - arranging \: both \: the \: equation \: in \: ax+$

$by+c=0 form: \\ </p><p>bx+ay−(a </p><p>2</p><p> +b </p><p>2</p><p> )=0 ....(3) \\ </p><p>ax−by+0=0 .....(4)$

$From \: equation \: (3) \: and \: (4) \: will \: get$

$a {}^{1} = \: b, \: b {}^{1} = a, \: c {}^{1} = - (a {}^{2} + b {}^{2} )$

$a {}^{2} = a,b² = -b, c² = 0$

$Now \: applying \: cross \: multiplication:$

$\frac{ \times }{b¹c² \: - \: b²c¹} = \frac{y}{c {}^{1}a² \: - c²a¹ } = \: \frac{1}{a¹b² \: - a²b¹ }$

$(b) \: \frac{ \times }{(0) -( - b) \: ( -a² - \: b²) } =$

$\frac{y}{(-a² \: - b²)a - 0 (b)} = \frac{1}{b(-b) \: - a(a)}$

$⇒ \frac{ \times }{0 +b(a² -b²) } = \frac{y}{(-a² - b²)a - 0} = \frac{1}{-a² -b² }$

$⇒ \frac{x}{b( -a² -b²) } = \frac{1}{ - a² - b² }$

$⇒x = b$

$⇒ \frac{y}{(-a²-b²)a} = \: \frac{1}{ - a²-b²}$

$⇒y = a$

Hence, x−y⇒b−a