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Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

                (vii)    \\\frac{10}{x + y} + \frac{2}{x - y}= 4\\ \frac{15}{x+y} - \frac{5}{x - y} = -2

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Given Equations,

\\\frac{10}{x + y} + \frac{2}{x - y}= 4\\ \frac{15}{x+y} - \frac{5}{x - y} = -2

Let, 

\frac{1}{x+y}=p\:and\:\frac{1}{x-y}=q

Now, our equation becomes

10p+2q=4........(1)

And

15p-5q=-2..........(2)

By Cross Multiplication method,

\frac{p}{b_1c_2-b_2c_1}=\frac{q}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{p}{(2)(2)-(-5)(-4)}=\frac{q}{(-4)(15)-(2)(10)}=\frac{1}{(10)(-5)-(15)(2)}

\frac{p}{4-20}=\frac{q}{-60-20}=\frac{1}{-50-30}

\frac{p}{-16}=\frac{q}{-80}=\frac{1}{-80}

p=\frac{1}{5},\:and\:q=1

Now,

p=\frac{1}{5}=\frac{1}{x+y}

\Rightarrow x+y=5........(3)

And,

q=1=\frac{1}{x-y}

\Rightarrow x-y=1...........(4)

Adding (3) and (4) we get,

\Rightarrow 2x=6

\Rightarrow x=3

Putting this value in (3) we get,

3+y=5

\Rightarrow y=2

And Hence,

x=3\:and\:y=2.

Posted by

Pankaj Sanodiya

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