Solve: ax – by = c; bx – ay = 1 + c.

Hint: Assume the given equations as equation (i) and equation (ii) respectively. To solve these equations, multiply equation (i) with ‘b’ and equation (ii) with ‘a’. Now, subtract the two equations obtained. This process will give the value of y. Now, substitute the value of y in one of the two provided equations to get the value of x. This process is known as the elimination method of solving linear equations.

Complete step-by-step answer:
We have been provided with the equations:
$\begin{align}
  & ax-by=c......................(i) \\
 & bx-ay=1+c.................(ii) \\
\end{align}$
Clearly, we can see that these are the linear equations in two variables. So, let us solve these equations by the help of elimination method. We have to eliminate one of the variables to find the other.
Now, multiplying equation (i) with b, we get,
$abx-{{b}^{2}}y=bc.....................(iii)$
And, multiplying equation (ii) with a, we get,
$abx-{{a}^{2}}y=a+ac.................(iv)$
Subtracting equation (iii) from equation (iv), we get,
$\begin{align}
  & \left( abx-{{a}^{2}}y \right)-\left( abx-{{b}^{2}}y \right)=a+ac-bc \\
 & \Rightarrow {{b}^{2}}y-{{a}^{2}}y=a+ac-bc \\
 & \Rightarrow y\left( {{b}^{2}}-{{a}^{2}} \right)=a+ac-bc \\
 & \Rightarrow y=\dfrac{a+ac-bc}{\left( {{b}^{2}}-{{a}^{2}} \right)} \\
\end{align}$
Now, substituting the value of y in equation (i), we get,
$\begin{align}
  & ax-b\times \left( \dfrac{a+ac-bc}{{{b}^{2}}-{{a}^{2}}} \right)=c \\
 & \Rightarrow ax=c+b\times \left( \dfrac{a+ac-bc}{{{b}^{2}}-{{a}^{2}}} \right) \\
 & \Rightarrow ax=c+\left( \dfrac{ab+abc-{{b}^{2}}c}{{{b}^{2}}-{{a}^{2}}} \right) \\
\end{align}$
Taking L.C.M in the R.H.S, we get,
\[\begin{align}
  & ax-b\times \left( \dfrac{a+ac-bc}{{{b}^{2}}-{{a}^{2}}} \right)=c \\
 & \Rightarrow ax=c+b\times \left( \dfrac{a+ac-bc}{{{b}^{2}}-{{a}^{2}}} \right) \\
 & \Rightarrow ax=\dfrac{{{b}^{2}}c-{{a}^{2}}c+\left( ab+abc-{{b}^{2}}c \right)}{{{b}^{2}}-{{a}^{2}}} \\
 & \Rightarrow ax=\dfrac{{{b}^{2}}c-{{a}^{2}}c+ab+abc-{{b}^{2}}c}{{{b}^{2}}-{{a}^{2}}} \\
 & \Rightarrow ax=\dfrac{-{{a}^{2}}c+ab+abc}{{{b}^{2}}-{{a}^{2}}} \\
\end{align}\]
Dividing both sides with ‘a’, we get,
\[\begin{align}
  & x=\dfrac{-ac+b+bc}{{{b}^{2}}-{{a}^{2}}} \\
 & \Rightarrow x=\dfrac{b+bc-ac}{{{b}^{2}}-{{a}^{2}}} \\
\end{align}\]

Note: One may note that we have solved this method by elimination method. We can also solve this system of equations with the help of a substitution method. In this method we have to substitute the value of one of the variables in terms of the other. In the above solution we can substitute the value of variable ‘y’ from equation (ii) in equation (i), to get the value of ‘x’.