How do you solve the linear equation system $\left( {\dfrac{4}{x}} \right) - \left( {\dfrac{3}{y}} \right) = 1$ & $\left( {\dfrac{6}{x}} \right) - \left( {\dfrac{3}{y}} \right) = - 4$?

Hint: First substitute $\dfrac{1}{x} = u$ and $\dfrac{1}{y} = v$ in the given system of equation. Next, multiply the equations so as to make the coefficients of the variables to be eliminated equal. Next, add or subtract the equation the equations obtained according as the terms having the same coefficients are of opposite or of the same sign. Solve the equation in one variable. Substitute the value found in any of the equations after substitution and find the value of another variable. Next, put the value of $u$ and $v$ in $\dfrac{1}{x} = u$ and $\dfrac{1}{y} = v$ respectively, and determine the value of $x$, and$y$. The value

Formula used:
Method of Elimination:
In this method, we eliminate one of the two variables to obtain an equation in one variable which can be easily solved. Putting the value of this variable in any one of the given equations, the value of another variable can be obtained.

Complete step by step answer:
The given system of equations is
$\left( {\dfrac{4}{x}} \right) - \left( {\dfrac{3}{y}} \right) = 1$
$\left( {\dfrac{6}{x}} \right) - \left( {\dfrac{3}{y}} \right) = - 4$
Taking $\dfrac{1}{x} = u$ and $\dfrac{1}{y} = v$. The given system of equations become
$4u - 3v = 1$…(i)
$6u - 3v = - 4$…(ii)
Now, multiply the equations so as to make the coefficients of the variables to be eliminated equal.
Here, the coefficient of the variable $v$ is already equal.
Next, add or subtract the equation the equations obtained according as the terms having the same coefficients are of opposite or of the same sign.
We can eliminate variable $v$ by subtracting equations (ii) from (i).
$ - 2u = 5$
Divide both sides of equation by $ - 2$, we get
$ \Rightarrow u = - \dfrac{5}{2}$
Now, substitute the value of $u$ in equation (i) and find the value of $v$.
$4\left( { - \dfrac{5}{2}} \right) - 3v = 1$
$ \Rightarrow - 10 - 3v = 1$
Add $10$ to both sides of the equation, we get
$ \Rightarrow - 3v = 11$
Divide both sides of equation by $ - 3$, we get
$ \Rightarrow v = - \dfrac{{11}}{3}$
Put the value of $u$ and $v$ in $\dfrac{1}{x} = u$ and $\dfrac{1}{y} = v$ respectively, and determine the value of $x$, and$y$.
$x = \dfrac{1}{u} = - \dfrac{2}{5}$ and $y = \dfrac{1}{v} = - \dfrac{3}{{11}}$

Hence, the solution of the given system of equation is $x = - \dfrac{2}{5}$, $y = - \dfrac{3}{{11}}$.

Note: We can also find the solution of a given system by Method of Cross-Multiplication.
System of equations:
$4u - 3v - 1 = 0$
$6u - 3v + 4 = 0$
By cross-multiplication, we have
$\dfrac{u}{{\begin{array}{*{20}{c}}
  { - 3}&{ - 1} \\
  { - 3}&4
\end{array}}} = \dfrac{{ - v}}{{\begin{array}{*{20}{c}}
  4&{ - 1} \\
  6&4
\end{array}}} = \dfrac{1}{{\begin{array}{*{20}{c}}
  4&{ - 3} \\
  6&{ - 3}
\end{array}}}$
$ \Rightarrow \dfrac{u}{{ - 12 - 3}} = \dfrac{{ - v}}{{16 + 6}} = \dfrac{1}{{ - 12 + 18}}$
$ \Rightarrow \dfrac{u}{{ - 15}} = \dfrac{{ - v}}{{22}} = \dfrac{1}{6}$
$ \Rightarrow u = \dfrac{{ - 15}}{6} = - \dfrac{5}{2}$ and $v = - \dfrac{{22}}{6} = - \dfrac{{11}}{3}$
$ \Rightarrow x = \dfrac{1}{u} = - \dfrac{2}{5}$ and $y = \dfrac{1}{v} = - \dfrac{3}{{11}}$
Final solution: Hence, the solution of the given system of equations is $x = - \dfrac{2}{5}$, $y = - \dfrac{3}{{11}}$.