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Solving by Elimination, Step by Step

This is how systems are actually solved — not by clever formulas, but by Gaussian elimination: three plain row operations that simplify the augmented matrix into a staircase shape, after which the answer falls out by back-substitution. We work one full 2x2 example all the way to the finish.

The three legal moves

Gaussian elimination works on the augmented matrix [A | b] using only three row operations. Each one changes how the system looks but never changes its solution — that is the whole reason they are allowed.

  1. Swap two rows (reorder the equations).
  2. Multiply a row by a nonzero number (scale an equation).
  3. Add a multiple of one row to another row (combine equations to kill a term).

Climbing to the staircase

The goal is row echelon form: a staircase where every row's leading number sits to the right of the one above it, so the bottom-left corner fills up with zeros. Once you reach it, the last row gives one unknown outright.

Let's take the system from the previous guide and write its augmented matrix, then clear the entry under the first pivot (the leading 2 in row 1).

[A|b] = [ 2  1 | 5 ]
        [ 1  3 | 6 ]

R2 -> R2 - (1/2)*R1:
        [ 2  1   | 5   ]
        [ 0  2.5 | 3.5 ]   <- staircase / row echelon form
One row operation clears the lower-left to 0; now it is a staircase.

Back-substitution finishes it

Now read the staircase from the bottom up. The last row says 2.5*y = 3.5, so y = 1.4. Substitute that into the first row, 2x + y = 5, and solve for x. This bottom-up unwinding is called back-substitution.

row 2:  2.5*y = 3.5  ->  y = 1.4
row 1:  2x + 1.4 = 5  ->  2x = 3.6  ->  x = 1.8

check:  1*1.8 + 3*1.4 = 1.8 + 4.2 = 6  (matches b)
Solve bottom row first, then back-substitute, then verify against b.