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## solve the following a(x+y)+b(x-y)-(a²-ab+b²)=0 and a(x+y)-b(x-y)-(a²+ab+b²)=0

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solve the following a(x+y)+b(x-y)-(a²-ab+b²)=0 and a(x+y)-b(x-y)-(a²+ab+b²)=0

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Mathematics
1 month
2021-08-06T08:01:58+00:00
2021-08-06T08:01:58+00:00 1 Answers
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## Answers ( )

Step-by-step explanation:Taking x+y=u and x−y=v the given system of equations becomes

au+bu−(a

2

−ab+b

2

)=0

au−bv−(a

2

+ab+b

2

)=0

By cross-multiplication, we have

⇒

b×−(a

2

+ab+b

2

)−(−b)×−(a

2

−ab+b

2

)

u

=

a×−(a

2

+ab+b

2

)+a(a

2

−ab+b

2

)

−v

=

a×−b−a×b

1

⇒

−b(a

2

+ab+b

2

)−−b(a

2

−ab+b

2

)

u

=

−a(a

2

+ab+b

2

)+a(a

2

−ab+b

2

)

−v

=

−ab−ab

1

⇒

−b(a

2

+ab+b

2

+a

2

−ab+b

2

)

u

=

−a(a

2

+ab+b

2

−a

2

+ab−b

2

)

−v

=

−2ab

1

⇒

−2b(a

2

+b

2

)

u

=

−a(2ab)

−v

=

−2ab

1

⇒u=

−2ab

−2b(a

2

+b

2

)

,v=

−2ab

2a

2

b

⇒u=

a

a

2

+b

2

,v=−a

Now, u=

a

a

2

+b

2

⇒x+y=

a

a

2

+b

2

.(i)

and, v=−a⇒x−y=−a ..(ii)

Adding equations (i) and (ii), we get

2x=

a

a

2

+b

2

−a⇒2x=

a

a

2

+b

2

−a

2

⇒2x=

a

b

2

⇒x=

2a

b

2

Substitutiing equation (ii) from equation (i), we get

2y=

a

a

2

+b

2

+a⇒2y=

a

a

2

+b

2

+a

2

⇒y=

2a

2a

2

+b

2

Hence, the solution of the given system of equations is x=

2a

b

2

,y=

2a

2a

2

+b

2

.