Math, asked by nehamehar1, 1 month ago

a/x-b/y=0
ab^2/x+a^2b/y=a^2+b^2​

Answers

Answered by Ameen1807
0

Answer

Taking  

x

1

=u and  

y

1

=v, the above system of equations becomes

au−bv+0=0

ab  

2

u+a  

2

bv−(a  

2

−b  

2

)=0

By cross-multiplication, we have

−b×−(a  

2

+b  

2

)−a  

2

b×0

u

=  

a×−(a  

2

+b  

2

)−ab  

2

×0

−v

=  

a×a  

2

b−ab  

2

×−b

1

 

⇒  

b(a  

2

+b  

2

)

u

=  

−a(a  

2

+b  

2

)

−v

=  

a  

3

b+ab  

3

 

1

 

⇒  

b(a  

2

+b  

2

)

u

=  

a(a  

2

+b  

2

)

v

=  

b(a  

2

+b  

2

)

1

 

⇒  

b(a  

2

+b  

2

)

u

=  

a(a  

2

+b  

2

)

v

=  

ab(a  

2

+b  

2

)

1

 

⇒u=  

ab(a  

2

+b  

2

)

b(a  

2

+b  

2

)

=  

a

1

 and  

v=  

ab(a  

2

+b  

2

)

a(a  

2

+b  

2

)

=  

b

1

 

Now, u=  

a

1

⇒  

x

1

=  

a

1

⇒x=a and v=  

b

1

⇒  

y

1

=  

b

1

⇒y=b

Hence, the solution of the given system of equation is x=a,y=b.

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