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

y \ne 0$.

Answer

Verified

362.7k+ views

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.

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.

Last updated date: 28th Sep 2023

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