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# Solve the following pairs of equations by reducing them to a pair of linear equations: (vii) Pair of Linear Equations in Two Variables Exercise 3.6 1

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,

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

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