Question

Solve the simultaneous equations :-

$x + y \div xy = 3 \\ x - y \div xy = 1$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm :\longmapsto\:\dfrac{x + y}{xy} = 3$

and

$\rm :\longmapsto\:\dfrac{x - y}{xy} = 1$

Above equations can be rewritten as

From first equation

$\rm :\longmapsto\:\dfrac{x + y}{xy} = 3$

$\rm :\longmapsto\:\dfrac{x}{xy} + \dfrac{y}{xy} = 3$

$\rm :\longmapsto\:\dfrac{1}{y} + \dfrac{1}{x} = 3 - - - (1)$

And from Second equation

$\rm :\longmapsto\:\dfrac{x - y}{xy} = 1$

$\rm :\longmapsto\:\dfrac{x}{xy} - \dfrac{y}{xy} = 1$

$\rm :\longmapsto\:\dfrac{1}{y} - \dfrac{1}{x} = 1 - - - - (2)$

Now, On adding equation (1) and (2), we get

$\rm :\longmapsto\:\dfrac{2}{y} = 3 + 1$

$\rm :\longmapsto\:\dfrac{2}{y} = 4$

$\rm :\longmapsto\:y = \dfrac{2}{4}$

$\bf\implies \:\:y = \dfrac{1}{2}$

Now, on Subtracting equation (2) from equation (1) we have

$\rm :\longmapsto\:\dfrac{2}{x} = 3 - 1$

$\rm :\longmapsto\:\dfrac{2}{x} = 2$

$\rm :\longmapsto\:x = \dfrac{2}{2}$

$\bf\implies \:x = 1$

Hence,

Solution of Simultaneous linear equations are

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x = 1} \\ \\ &\sf{y = \dfrac{1}{2} } \end{cases}\end{gathered}\end{gathered}$

Answer 2

Answer:

your solution is as follows

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Step-by-step explanation:

$to \:solve \: for : \\ values \: of \: x \: and \: y \\ \\ given \: linear \: equations \: are \\ \frac{x + y}{xy} = 3 \: \: \: \: \: \: \: (1) \\ \\ \frac{x - y}{xy} = 1 \: \: \: \: \: \: (2) \\ \\ applying \: now \: \: \frac{(1)}{(2)} \\ we \: get \\ \\ \frac{x + y}{x - y} = \frac{3}{1} \\ \\ using \: componendo \: dividendo \\ we \: get \\ \\ \frac{x + y + x - y}{x + y - (x - y)} = \frac{3 + 1}{3 - 1} \\ \\ \frac{x + x}{y + y} = \frac{4}{2} \\ \\ \frac{2x}{2y} = 2 \\ \\ x = 2y \: \: \: \: \: \: \: \: \: \: (3)$

$substituting \: value \: of \: x \: in \: (1) \\ we \: have \\ \\ \frac{x + y}{xy} = 3 \\ \\ \frac{2y + y}{xy} = 3 \\ \\ \frac{3y}{xy} = 3 \\ \\ \frac{1}{x} = 1 \\ \\ x = 1 \\ \\ on \: substituting \: value \: of \: x \: in \: (3) \\ we \: have \\ \\ y = \frac{1}{2}$

$hence \: here \: \\ (x,y) = (1, \frac{1}{2} \: )$