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Q7.    Solve the following pair of linear equations:

                (iv)    \\(a-b)x + (a+b)y = a^2 -2ab - b^2\\ (a+b)(x+y) = a^2 +b^2

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Given,

\\(a-b)x + (a+b)y = a^2 -2ab - b^2..........(1)

And

\\ (a+b)(x+y) = a^2 +b^2\\\Rightarrow (a+b)x+(a+b)y=a^2+b^2...........(2)

Now, Subtracting (1) from (2), we get

(a+b)x-(a-b)x=a^2+b^2-a^2+2ab+b^2

\Rightarrow(a+b-a+b)x=2b^2+2ab

\Rightarrow 2bx=2b(b+2a)

\Rightarrow x=(a+b)

Substituting this in (1), we get,

(a-b)(a+b)+(a+b)y=a^2-2ab-b^2

\Rightarrow a^2-b^2+(a+b)y=a^2-2ab-b^2

\Rightarrow (a+b)y=-2ab

\Rightarrow y=\frac{-2ab}{a+b}.

Hence,

x=(a+b),\:and\:y=\frac{-2ab}{a+b}

 

 

 

Posted by

Pankaj Sanodiya

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