Substitution and elimination both make you rearrange the equations. Cramer's method does not. You write down three determinants, divide, and the answers fall out. Every step is fixed, which is why it turns into an algorithm a computer can run.

Concept The coefficient matrix and its determinant

Read the coefficients straight off the system, keeping every negative sign. For $5x - 6y = 15$ and $3x + 4y = -29$ :

D = \begin{vmatrix} 5 & -6 \\ 3 & 4 \end{vmatrix} = (5)(4) - (-6)(3) = 20 + 18 = 38

A $2 \times 2$ determinant is the first diagonal minus the second diagonal. Compute $D$ first: it is the denominator for every variable, and if it comes out $0$ the method cannot be used.

Concept Replace a column to get its determinant

The denominator is always

D

. For

D_x

, remove the

x

-column and drop the constants in its place. For

D_y

, remove the

y

-column instead. Nothing else changes.

x = \frac{D_x}{D} \qquad y = \frac{D_y}{D}

Example The full run

Solve $5x - 6y = 15$ and $3x + 4y = -29$ .

D = (5)(4) - (-6)(3) = 38

D_x = (15)(4) - (-6)(-29) = 60 - 174 = -114

D_y = (5)(-29) - (15)(3) = -145 - 45 = -190

x = \dfrac{-114}{38} = -3 \qquad y = \dfrac{-190}{38} = -5

⟹ x = −3, y = −5

Watch the sign in $D_x$ : subtracting $(-6)(-29)$ removes a positive product, which is what pulls the value down to $-114$ .

Note Why it is called systematic

Nothing is rearranged at any point — the equations are only ever read, never manipulated.

The same three operations run in the same order for every system.

That fixed pattern is exactly what a computer needs, which is why linear solvers use it.

The one thing to check is

D \neq 0

; if

D = 0

there is no unique solution.

Summary

Build the coefficient matrix from the system, keeping the negative signs.
Compute $D$ first; it is the denominator for every variable and must not be $0$ .
For $D_x$ replace the $x$ -column with the constants; for $D_y$ replace the $y$ -column.
$x = D_x / D$ and $y = D_y / D$ .
For $5x - 6y = 15$ , $3x + 4y = -29$ : $D = 38$ , $D_x = -114$ , $D_y = -190$ , so $x = -3$ , $y = -5$ .

Cramer's Method