Question

ax+by=1 and bx+ay=2ab/a2+b2
find x and y​

Answer 1

Given :

•The given equation are

$\bullet \bf \: ax + by = 1 \: \: \: \: ...(1)$

$\\ \bullet \bf \: bx + ay = \frac{2ab}{ {a}^{2} + {b}^{2} } \: \: ....(2)$

To Find :

•The value of x and y =?

Solution :

The given equation are :

$\bf \bullet \: ax + by = 1 \: \: ...(1)$

$\\ \bullet \bf \: bx + ay = \frac{2ab}{ {a}^{2} + {b}^{2} }$

By cross multiplication rule, we get

$\\ \implies \sf \: \frac{x}{ \frac{ - 2a {b}^{2} +a }{ {a}^{2} + {b}^{2} } } = \frac{y}{ - b + \frac{2 {a}^{2} b}{ {a}^{2} + {b}^{2} } } = \frac{1}{ {a}^{2} - {b}^{2} }$

$\\ \implies \sf \: \frac{x}{ \frac{ - 2a {b}^{2} + {a}^{3} + a {b}^{2} }{ {a}^{2} + {b}^{2} } } = \frac{y}{ \frac{ - {a}^{2}b - {b}^{3} + 2 {a}^{2}b }{ {a}^{2} + {b}^{2} } } = \frac{1}{ {a}^{2} - {b}^{2} }$

$\\ \implies \sf \: \frac{x}{ \frac{ {a}^{3} - a{b}^{2} }{ {a}^{2} + {b}^{2} } } = \frac{y}{ \frac{ {a}^{2}b - {b}^{3} }{ {a}^{2} + {b}^{2} } } = \frac{1}{ {a}^{2} - {b}^{2} }$

$\\ \implies \sf \: \frac{x}{ \frac{a( {a}^{2} - {b}^{2} ) }{ {a}^{2} + {b}^{2} } } = \frac{y}{ \frac{b( {a}^{2} - {b}^{2}) }{ {a}^{2} + {b}^{2} } } = \frac{1}{ {a}^{2} - {b}^{2} }$

$\\ \implies \sf \: \frac{x}{ \frac{a}{ {a}^{2} + {b}^{2} } } = 1$

$\implies \boxed{ \bf{ x = \frac{a}{ {a}^{2} + {b}^{2} } }}$

And,

$\\ \implies \sf \: \frac{y}{ \frac{b}{ {a}^{2} + {b}^{2} } } = 1$

$\implies \boxed{ \bf{y = \frac{b}{ {a}^{2} + {b}^{2} } }}$

The value of x and y is :

$\\ \implies \boxed{ \sf{x = \frac{a}{ {a}^{2} + {b}^{2} } \: \: and \: \: y = \frac{b}{ {a}^{2} + {b}^{2} } }}$

Answer 2

Step-by-step explanation:

Given :

•The given equation are

\bullet \bf \: ax + by = 1 \: \: \: \: ...(1)∙ax+by=1...(1)

\begin{gathered} \\ \bullet \bf \: bx + ay = \frac{2ab}{ {a}^{2} + {b}^{2} } \: \: ....(2)\end{gathered}

∙bx+ay=

a

2

+b

2

2ab

....(2)

To Find :

•The value of x and y =?

Solution :

The given equation are :

\bf \bullet \: ax + by = 1 \: \: ...(1)∙ax+by=1...(1)

\begin{gathered} \\ \bullet \bf \: bx + ay = \frac{2ab}{ {a}^{2} + {b}^{2} } \end{gathered}

∙bx+ay=

a

2

+b

2

2ab

By cross multiplication rule, we get

\begin{gathered} \\ \implies \sf \: \frac{x}{ \frac{ - 2a {b}^{2} +a }{ {a}^{2} + {b}^{2} } } = \frac{y}{ - b + \frac{2 {a}^{2} b}{ {a}^{2} + {b}^{2} } } = \frac{1}{ {a}^{2} - {b}^{2} } \end{gathered}

⟹

a

2

+b

2

−2ab

2

+a

x

=

−b+

a

2

+b

2

2a

2

b

y

=

a

2

−b

2

1

\begin{gathered} \\ \implies \sf \: \frac{x}{ \frac{ - 2a {b}^{2} + {a}^{3} + a {b}^{2} }{ {a}^{2} + {b}^{2} } } = \frac{y}{ \frac{ - {a}^{2}b - {b}^{3} + 2 {a}^{2}b }{ {a}^{2} + {b}^{2} } } = \frac{1}{ {a}^{2} - {b}^{2} } \end{gathered}

⟹

a

2

+b

2

−2ab

2

+a

3

+ab

2

x

=

a

2

+b

2

−a

2

b−b

3

+2a

2

b

y

=

a

2

−b

2

1

\begin{gathered} \\ \implies \sf \: \frac{x}{ \frac{ {a}^{3} - a{b}^{2} }{ {a}^{2} + {b}^{2} } } = \frac{y}{ \frac{ {a}^{2}b - {b}^{3} }{ {a}^{2} + {b}^{2} } } = \frac{1}{ {a}^{2} - {b}^{2} } \end{gathered}

⟹

a

2

+b

2

a

3

−ab

2

x

=

a

2

+b

2

a

2

b−b

3

y

=

a

2

−b

2

1

\begin{gathered} \\ \implies \sf \: \frac{x}{ \frac{a( {a}^{2} - {b}^{2} ) }{ {a}^{2} + {b}^{2} } } = \frac{y}{ \frac{b( {a}^{2} - {b}^{2}) }{ {a}^{2} + {b}^{2} } } = \frac{1}{ {a}^{2} - {b}^{2} } \end{gathered}

⟹

a

2

+b

2

a(a

2

−b

2

)

x

=

a

2

+b

2

b(a

2

−b

2

)

y

=

a

2

−b

2

1

\begin{gathered} \\ \implies \sf \: \frac{x}{ \frac{a}{ {a}^{2} + {b}^{2} } } = 1\end{gathered}

⟹

a

2

+b

2

a

x

=1

\implies \boxed{ \bf{ x = \frac{a}{ {a}^{2} + {b}^{2} } }}⟹

x=

a

2

+b

2

a

And,

\begin{gathered} \\ \implies \sf \: \frac{y}{ \frac{b}{ {a}^{2} + {b}^{2} } } = 1\end{gathered}

⟹

a

2

+b

2

b

y

=1

\implies \boxed{ \bf{y = \frac{b}{ {a}^{2} + {b}^{2} } }}⟹

y=

a

2

+b

2

b

The value of x and y is :

[tex] \begin{gathered} \\ \implies \boxed{ \sf{x = \frac{a}{ {a}^{2} + {b}^{2} } \: \: and \: \: y = \frac{b}{ {a}^{2} + {b}^{2} } }}\end{gathered} [\tex]

⟹

x=

a

2

+b

2

a

andy=

a

2

+b

2

b. pls report this answer I wrote it mistakely.