Question

Solve the following systems of equations: $\dfrac{{xy}}{{x + y}} = \dfrac{6}{5}$, \[\dfrac{{xy}}{{y - x}} = 6\] where \[x + y \ne 0\] and \[y - x \ne 0\].

Answer 1

Hint: In these types of questions simplify the equations$\dfrac{{xy}}{{x + y}} = \dfrac{6}{5}$, \[\dfrac{{xy}}{{y - x}} = 6\], and divide the equation by xy and then find the value of x and y.

Complete step-by-step answer:

Given equation: $\dfrac{{xy}}{{x + y}} = \dfrac{6}{5}$, \[\dfrac{{xy}}{{y - x}} = 6\]

$\dfrac{{xy}}{{x + y}} = \dfrac{6}{5}$
$ \Rightarrow $$6x + 6y = 5xy$ (Equation 1)
\[\dfrac{{xy}}{{y - x}} = 6\]
$ \Rightarrow $\[6y - 6x = xy\] (Equation 2)
Dividing equation 1 and equation 2 by xy
$ \Rightarrow $\[\dfrac{{6x}}{{xy}} + \dfrac{{6y}}{{xy}} = \dfrac{{5xy}}{{xy}}\]
$ \Rightarrow $\[\dfrac{6}{y} + \dfrac{6}{x} = 5\] (Equation 3)
$ \Rightarrow $\[\dfrac{{6y}}{{xy}} - \dfrac{{6x}}{{xy}} = \dfrac{{xy}}{{xy}}\]
$ \Rightarrow $\[\dfrac{6}{x} - \dfrac{6}{y} = 1\] (Equation 4)
Adding equation 3 and equation 4
$ \Rightarrow $\[\dfrac{6}{y} + \dfrac{6}{x} + \dfrac{6}{x} - \dfrac{6}{y} = 1 + 5\]
$ \Rightarrow $\[\dfrac{{12}}{x} = 6\]
$ \Rightarrow $\[x = 2\]
Putting the value of x in equation 3
\[\dfrac{6}{y} + \dfrac{6}{2} = 5\]
$ \Rightarrow $\[y = 3\]
Hence values of x and y are 2, 3.

Note: In these types of questions use the given equation i.e.$\dfrac{{xy}}{{x + y}} = \dfrac{6}{5}$, \[\dfrac{{xy}}{{y - x}} = 6\] and simplify it and then divide the equation 1 i.e. $6x + 6y = 5xy$ and equation 2 i.e. \[6y - 6x = xy\] by xy and then add both the equation and find the value of x and then use the value of x to find the value of y.