Substitution and elimination

Graphing shows that two lines cross; algebra finds where, exactly. Two methods do it, and both work by the same trick: turn two unknowns into one.

Substitution: replace and solve

Best when an equation already has a variable alone (or almost alone).

$\begin{cases} y = 2x - 1 \\ 3x + y = 9 \end{cases}$

The first equation tells you what y is. So wherever y appears in the second, write 2x − 1 instead:

Solution: (2, 3). Check in both originals: 3 = 2(2)−1 ✓ and 3(2)+3 = 9 ✓.

Elimination: add the equations

Best when the equations are lined up in ax + by = c form.

$\begin{cases} 2x + 3y = 12 \\ 5x - 3y = 9 \end{cases}$

The y-coefficients are +3 and −3 — opposites. Add the equations column by column and y cancels: 7x = 21, so x = 3, and then y = 2 from either equation. Solution: (3, 2).

If nothing cancels on its own, make it cancel: to solve 3x + 2y = 7 with 2x + 5y = 12, multiply the first by 2 and the second by −3 (giving +6x and −6x), then add. Multiplying a whole equation by a number doesn't change its line — it's the balance rule again.

Choosing in one glance

When the variables all vanish

Both methods can end with no variables left. A true leftover (0 = 0) means the lines were identical — infinitely many solutions. A false one (0 = 7) means parallel — none. Same three fates as the picture from last lesson, discovered algebraically.

Why this matters

Systems are the first time algebra handles several facts at once — and the technique of eliminating variables scales far beyond two: the same idea, as matrix row reduction, is how computers solve systems with thousands of unknowns.