Question

If bx + ay = a^2 + b^2 and ax - by = 0, then the value of x - y equals:
(a) a - b
(b) b - a
(c) a^2 - b^2
(d) b^2 + a²

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

If bx + ay = a² + b² and ax - by = 0, then the value of x - y equals:

(a) a - b

(b) b - a

(c) a^2 - b^2

(d) b^2 + a²

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{\pink{Given}}}}}$

• bx + ay = a² + b² ............(1)
• ax - by = 0 ................….(2)

$\Large{\underline{\mathfrak{\bf{\pink{Find}}}}}$

• Value of (x - y )

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

multiply by "a" in equ(1) and " b" in equ(2)

• ab x + a² y = a³ + ab²
• ab x - b² y = 0

________________Sub. its,

$\mapsto\sf{\:y(a^2+b^2)\:=\:(a^3+ab^2)} \\ \\ \mapsto\sf{\:y\:=\:\dfrac{a\cancel{(a^2+b^2)}}{\cancel{(a^2+b^2)}}} \\ \\ \mapsto\sf{\orange{\:y\:=\:a}}$

keep value of "y" in equ(2),

$\mapsto\sf{\:(ax-ba)\:=\:0} \\ \\ \mapsto\sf{\:ax\:=\:ba} \\ \\ \mapsto\sf{\:x\:=\:\dfrac{b\cancel{a}}{\cancel{a}}} \\ \\ \mapsto\sf{\orange{\:x\:=\:b}}$

$\Large{\underline{\mathfrak{\bf{\blue{Thus}}}}}$

• Value of x = b
• Value of y = a

Now,

$\mapsto\sf{\green{\:(x-y)}} \\ \\ \small\sf{\:\:\:\:(Keep\:Value\:of\:x\:and\:y)} \\ \\ \mapsto\bf{\:(b-a)}$

Hence,

$\sf{\green{\:(Answer\:will\:be\:option\:number\:(a)}}$

$\Large{\underline{\underline{\mathfrak{\bf{Answer\:Verification}}}}}$

keep value of "x" and "y" in equ(1)

$\mapsto\sf{\:b.(b)+a.(a)\:=\:(a^2+b^2)} \\ \\ \mapsto\sf{\:(a^2+b^2)\:=\:(a^2+b^2)} \\ \\ \bf{\:\:\:\:L.H.S.\:=\:R.H.S.}$

That's prived.

Hence, we can say that our solution is absolutely right .

Answer 2

Step-by-step explanation:

$\huge\underline\bold{Question-} \\ If \: bx + ay = a {}^{2} - b {}^{2} \: and \: ax - by = 0, \\ then \: find \: the \: value \: of \: x - y \: equals - \\ (a) a - b \\ (b) b - a \: \: \bold{(Correct)} \\ (c) a {}^{2} - b {}^{2} \\ (d) b {}^{2} + a {}^{2}$

$\huge\underline\bold{Solution-}$

$\bold{Given -} \\ bx + ay = {a}^{2} + {b}^{2}.....(a) \\ ax - by = 0 \: .....(b) \\ \\ \bold{To \: find - } \\ x - y = \: ? \\ \\ Now, \\ ax = by \: .... \bold{from \: (b)} \\ \bold{y = \frac{ax}{b} } \: \bold{or} \: \bold{x = \frac{by}{a} } \\ \\ \bold{Substituting \: the \: value \: of \: y \: on \: (a)} \\ \\ bx + a( \frac{ax}{b} ) = {a}^{2} + {b}^{2} \\ = bx + \frac{ {a}^{2}x }{b} = {a}^{2} + {b}^{2} \\ = \frac{ {b}^{2}x + {a}^{2}x }{b} = {a}^{2} + {b}^{2} \\ = x( {a}^{2} + {b}^{2} ) = {a}^{2} b + {b}^{ 3} \\ = \bold{x = \frac{ {a}^{2}b + {b}^{3} }{ {a}^{2} + {b}^{2} } }$ $Now, \\ \bold {Substituting \: the \: value \: of \: x \: on \: (a)} \\ b( \frac{by}{a} ) + ay = {a}^{2} + {b}^{2} \\ = \frac{ {b}^{2}y }{a} + ay = {a}^{2} + {b}^{2} \\ = \frac{ {b}^{2}y + {a}^{2}y }{a} = {a}^{2} + {b}^{2} \\ = y( {a}^{2} + {b}^{2} ) = {a}^{3} + a{b}^{2} \\ = \bold{y = \frac{ {a}^{3} + a {b}^{2} }{ {a}^{2} + {b}^{2} } }$ $Now, \\ x - y = \\ \\ = \frac{ {a}^{2}b + {b}^{3} }{ {a}^{2} + {b}^{2} } - \frac{ {a}^{3} + a {b}^{2} }{ {a}^{2} + {b}^{2} } \\ = \frac{ {a}^{2}b + {b}^{3} - {a}^{3} - a {b}^{2} }{ {a}^{2} + {b}^{2} } \\ = \frac{b( {a}^{2} + {b}^{2} ) - a( {a}^{2} + {b}^{2}) }{ {a}^{2} + {b}^{2} } \\ = \frac{ \cancel{( {a}^{2} + {b}^{2} )}(b - a)}{ \cancel{( {a}^{2} + {b}^{2} )}} \\ = b - a \\ \\ Hence, \\ \bold{The \: value \: of \: x - y \: is \: b - a}$