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Solve the following system of equations $x + y = 5xy$, $3x + 2y = 13xy,$ where $x \ne 0,
y \ne 0$.

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Hint: To solve this problem we need to convert the given equation into a proper equation where the R.H.S part should be constant. Here the proper equation can be obtained by dividing the equation with xy term on both sides.

Complete step-by-step answer:

Given equation are
$x + y = 5xy - - - - - - - - - > (1)$
$3x + 2y = 13xy - - - - - - - - > (2)$

Now here let us divide both the equation with $'xy'$ term then we get
From equation (1)
$ \Rightarrow \dfrac{x}{{xy}} + \dfrac{y}{{xy}} = \dfrac{{5xy}}{{xy}}$
On cancellation we get the equation as
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = 5 - - - - - - > (3)$
From equation (2)
$ \Rightarrow \dfrac{{3x}}{{xy}} + \dfrac{{2y}}{{xy}} = \dfrac{{13}}{{xy}}$
On cancellation we get
$ \Rightarrow \dfrac{3}{x} + \dfrac{2}{y} = 13 - - - - - - - - > (4)$
Now let us multiply equation (3) with 5 we get
$ \Rightarrow \dfrac{3}{x} + \dfrac{3}{y} = 15 - - - - - - - - > (5)$
On subtracting equations (5)(4) we get
$
   \Rightarrow \dfrac{1}{y} = 2 \\
   \Rightarrow y = \dfrac{1}{2} \\
 $

On substituting y value either equation (4) or (5) we get
$ \Rightarrow x = \dfrac{1}{3}$
Hence we solved both equations and got x,y values.

Note: In this problem to get proper equation format we have divided both the equation with xy term and later subtracted the equation .Generally we ignore to convert the given equation.
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