Solve for x and y:
 $ ax+by-2a+3b=0 $
 $ bx-ay=3a+2b $

Hint: Check for the consistency of the pair of equations by comparing the ratios of the coefficients of a and y in both the equations (see Note below).
A pair of simultaneous linear equations in two variables can be solved by either Cramer's rule (see Note below), by elimination, or by substitution.
The given system of equations can be solved quickly by eliminating one of the variables. In order to eliminate one of the variables, make its coefficient equal in both the equations by multiplying the equations by a proper number and then add/subtract the equations.

Complete step-by-step answer:
Rearranging the terms so as to keep only the variables on one side, the given equations are:
 $ ax+by=2a-3b $ ... (1)
 $ bx-ay=3a+2b $ ... (2)
We find that the ratio of the coefficients of $ x $ is $ a:b $ and the ratio of the coefficients of $ y $ is $ b:-a $ . Since, the ratios are not equal, the given system of equations has a unique solution (see Note below).
To find the solution, let us eliminate $ y $ by making its coefficients equal in both the equations.
Multiplying the equation (1) by $ a $ , and equation (2) by $ b $ , we get:
 $ {{a}^{2}}x+aby=2{{a}^{2}}-3ab $ ... (3)
 $ {{b}^{2}}x-aby=3ab+2{{b}^{2}} $ ... (4)
Adding the equations (3) and (4), we get:
 $ {{a}^{2}}x+{{b}^{2}}x=2{{a}^{2}}+2{{b}^{2}} $
Separating the common factors, we can write:
⇒ $ x\left( {{a}^{2}}+{{b}^{2}} \right)=2\left( {{a}^{2}}+{{b}^{2}} \right) $
Assuming that $ a,b\ne 0 $ , $ \left( {{a}^{2}}+{{b}^{2}} \right) $ is always positive (why?).
Therefore, on dividing both sides by $ \left( {{a}^{2}}+{{b}^{2}} \right) $ , we get:
⇒ $ x=2 $
Substituting this value of $ x $ in equation (1) [or equation (2)], we get:
 $ 2a+by=2a-3b $
Subtracting $ 2a $ from both the sides and dividing by $ b $ , we get:
⇒ $ by=-3b $
Dividing both sides by $ b $ , we get:
⇒ $ y=-3 $

Therefore, the solution of the given set of equations is $ x=2 $ and $ y=-3 $.

Note: The consistency of a pair of equations $ {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} $ and $ {{a}_{2}}x+{{b}_{2}}y={{c}_{2}} $ , is determined as follows:
 $ \dfrac{{{a}_{1}}}{{{a}_{2}}}\ne \dfrac{{{b}_{1}}}{{{b}_{2}}} $ : The system is consistent and has a unique solution.
The pair of equations represent a pair of intersecting lines.
 $ \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}} $ : The system is consistent and has infinitely many solutions.
The pair of equations represent the same line.
 $ \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}\ne \dfrac{{{c}_{1}}}{{{c}_{2}}} $ : The system is inconsistent, i.e. it has no solutions.
The pair of equations represent a pair of parallel lines.
The solutions to the equations $ {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} $ and $ {{a}_{2}}x+{{b}_{2}}y={{c}_{2}} $ can be determined using Cramer's rule:
 $ D=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} \\ {{b}_{1}} & {{b}_{2}} \\ \end{matrix} \right| $ , $ {{D}_{x}}=\left| \begin{matrix} {{c}_{1}} & {{c}_{2}} \\ {{b}_{1}} & {{b}_{2}} \\ \end{matrix} \right| $ , $ {{D}_{y}}=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} \\ {{c}_{1}} & {{c}_{2}} \\ \end{matrix} \right| $ .
The values of the variables x and y are given by: $ x=\dfrac{{{D}_{x}}}{D} $ and $ y=\dfrac{{{D}_{y}}}{D} $ .
If, D = 0, the system is:
Either consistent and has infinitely many solutions. In this case $ {{D}_{x}}={{D}_{y}}=0 $ .
Or it is inconsistent, i.e. it has no solutions. In this case $ {{D}_{x}}\ne 0 $ and $ {{D}_{y}}\ne 0 $ .