Q.The sum of three numbers is 15. If the second number is subtracted from the sum of first and third numbers then we get 5. When the third number is subtracted from the sum of twice the first number and the second number, we get 4. Find the three numbers.

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. …