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Mathematics

Solve :

3x+y+2xy=2\dfrac{3}{x+y} + \dfrac{2}{x-y} = 2 and

9x+y4xy=1\dfrac{9}{x+y} - \dfrac{4}{x-y} = 1

Linear Equations

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Answer

Let 1x+y\dfrac{1}{x + y} = u and 1xy\dfrac{1}{x - y} = v.

The given equations are :

3u + 2v = 2

9u - 4v = 1

Multiply the first equation by 3 and subtract the second equation from it.

⇒ (3u + 2v = 2) x 3

9u+6v=69u4v=1+10v=6110v=5\begin{matrix} & 9u & + & 6v & = & 6 \ & 9u & - & 4v & = & 1 \ & - & + & & & - \ \hline & & & 10v & = & 6 - 1\ \Rightarrow & & & 10v & = & 5 \end{matrix}

⇒ v = 510\dfrac{5}{10}

⇒ v = 12\dfrac{1}{2}

Substituting the value of v in first equation, we get:

⇒ 3u + 2 ×12\times \dfrac{1}{2} = 2

⇒ 3u + 1 = 2

⇒ 3u = 2 - 1

⇒ 3u = 1

⇒ u = 13\dfrac{1}{3}

So, x + y = 1u\dfrac{1}{u} = 3 and x - y = 1v\dfrac{1}{v} = 2

⇒ x + y = 3

And, x - y = 2

Adding both the equations, we get:

x+y=3xy=22x=3+22x=5\begin{matrix} & x & + & y & = & 3 \ & x & - & y & = & 2 \ \hline & 2x & & & = & 3 + 2\ \Rightarrow & 2x & & & = & 5 \end{matrix}

⇒ x = 52\dfrac{5}{2}

Substituting the value of x in first equation, we get:

52\dfrac{5}{2} + y = 3

⇒ y = 3 - 52\dfrac{5}{2}

⇒ y = 652\dfrac{6 - 5}{2}

⇒ y = 12\dfrac{1}{2}

Hence, the value of x = 52\dfrac{5}{2} and y = 12\dfrac{1}{2}.

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