Q.If $ae^x + be^y = c$; $pe^x + qe^y = d$ and $\Delta_1 = \begin{vmatrix}a & b\\p & q\end{vmatrix}$; $\Delta_2 = \begin{vmatrix}c & b\\d & q\end{vmatrix}$; $\Delta_3 = \begin{vmatrix}a & c\\p & d\end{vmatrix}$ then the value of $(x, y)$ is :

For a system of 3 linear equations in 3 unknowns $a_{i1}x_1+a_{i2}x_2+a_{i3}x_3=b_i$ ($i=1,2,3$) whose coefficient determinant $\Delta=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}$ is non-zero, Cramer's rule gives each unknown directly as a ratio of two determinants:

where $\Delta_k$ is $\Delta$ with its $k$th column replaced by the constants column $(b_1,b_2,b_3)^T$, everything else unchanged. (The same pattern extends to 2 equations in 2 unknowns: $\Delta=\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}$, $x=\Delta_1/\Delta$, $y=\Delta_2/\Delta$.)

Why it works. Multiplying $\Delta$ by $x_1$ and using the linearity-in-a-column property of determinants (splitting the first column $a_{i1}x_1$ into the sum $a_{i1}x_1+a_{i2}x_2+a_{i3}x_3$ using the original equations, then subtracting off the $x_2,x_3$ multiples of the identical columns 2 and 3, which vanish) collapses the first column to exactly the constants $b_i$ -- giving $x_1\Delta=\Delta_1$, and dividing by $\Delta\ne0$ gives the rule.

Worked illustration. For $x+y=3,\ 2x-y=0$: $\Delta=\begin{vmatrix}1&1\\2&-1\end{vmatrix}=-3$, $\Delta_1=\begin{vmatrix}3&1\\0&-1\end{vmatrix}=-3$, $\Delta_2=\begin{vmatrix}1&3\\2&0\end{vmatrix}=-6$. So $x=\Delta_1/\Delta=1$, $y=\Delta_2/\Delta=2$ -- check: $1+2=3$ and $2(1)-2=0$, correct.

Word problems that produce equations like $y=ax^2+bx+c$ through three given points, or rate/mixture/scoring problems, translate to a $3\times3$ system in the unknown constants exactly as for matrix inversion, then Cramer's rule reads off each unknown independently -- convenient when only one or two of the unknowns are actually needed. A system with fractional unknowns like $\dfrac{a}{x}+by=c$ is first turned linear by the substitution $u=\dfrac1x$ (or $\dfrac1y$, $\dfrac1z$), solved for $u,v,(w)$ by Cramer's rule, and only inverted back to $x,y,(z)$ at the very last step. …