Solving by Elimination

Two coffees and one tea cost \pounds 7. One coffee and one tea cost \pounds 5. What's the price of a coffee, and what's the price of a tea? You can't answer from either fact alone — but put them together and the prices are pinned down exactly. That is the whole business of simultaneous equations: two facts, two unknowns, one shared answer.

Look at those two facts as equations, with c for the price of a coffee and t for a tea:

2c + t = 7 \qquad c + t = 5

Both drinks have exactly one tea in them — a +t in each equation. So if you subtract the second from the first, the tea cancels and you're left with the difference: one extra coffee costs \pounds 2.

(2c + t) - (c + t) = 7 - 5 \;\Rightarrow\; c = 2

A coffee is \pounds 2. Slot that back into c + t = 5 and 2 + t = 5, so a tea is \pounds 3. That is elimination: combine the two equations so one unknown vanishes, leaving an ordinary equation you already know how to solve.

The method, step by step

Elimination always follows the same four moves. The clever part is only the first one — lining a variable up so it cancels.

  1. Match the coefficients of one variable in the two equations (multiply an equation through if you need to).
  2. Add or subtract the equations to make that variable disappear.
  3. Solve the single-variable equation that's left.
  4. Back-substitute that value into either original equation to find the other variable.

You're always allowed to add two true equations, or subtract one from the other — the result is still true. Choose whichever makes a variable drop out. Here is the classic pair where the y terms are already equal and opposite:

2x + y = 7 \qquad x - y = 2

The +y and -y are opposite, so adding wipes y out:

3x = 9 \;\Rightarrow\; x = 3

Then back-substitute x = 3 into x - y = 2: 3 - y = 2, so y = 1. The solution is x = 3,\ y = 1 — and it fits both equations, just as you'd check after .

See it built

Step through the cancellation. The two equations stack up; their y terms are equal and opposite, so adding them wipes y out and leaves a single equation in x. Solve that, then feed x back to find y.

When no pair of terms is already equal and opposite, first scale an equation — multiply it through by a number — so that a variable lines up to cancel. That is the same legal move as : whatever you do to one side, do to the other.

When the coefficients don't match — scale first

Most systems aren't so tidy. Nothing cancels straight away here:

3x + 2y = 16 \qquad x - y = 3

The y terms are +2y and -y — not a match. Double the second equation: x - y = 3 becomes 2x - 2y = 6. Now watch a cleaner pair instead so the arithmetic stays tidy. Take 3x + 2y = 13 and x - y = 1. Double the second to get 2x - 2y = 2. Now the y terms are +2y and -2y — opposite signs, so add:

5x = 15 \;\Rightarrow\; x = 3, \quad\text{then}\quad 3 - y = 1 \;\Rightarrow\; y = 2.

You could just as well have matched the x terms instead — multiply the second by 3 to get 3x - 3y = 3, then subtract from the first to kill x. Either variable works; pick whichever needs less arithmetic.

A word problem: cinema tickets

A family buys 2 adult and 3 child tickets for \pounds 32. Their neighbours buy 2 adult and 1 child ticket for \pounds 20. What does each ticket cost? Let a be the adult price and c the child price:

2a + 3c = 32 \qquad 2a + c = 20

Both have 2a — the same sign, the same coefficient. So subtract the second from the first and the adults cancel:

2c = 12 \;\Rightarrow\; c = 6.

A child ticket is \pounds 6. Back-substitute into 2a + c = 20: 2a + 6 = 20, so 2a = 14 and a = 7. Adults \pounds 7, children \pounds 6 — and it checks: 2(7) + 3(6) = 14 + 18 = 32. That's the same recipe as the coffee and tea, dressed in a new story.

The single most common slip is adding when you should subtract, or the reverse. Look at the signs of the matched coefficients:

Get it backwards and you don't cancel — you get a bigger term (adding +2y and +2y gives 4y, not zero). And when you subtract, subtract every term, right-hand side included. Subtracting x - y = 2 from 2x + y = 7 is (2x - x) + (y - (-y)) = 7 - 2, i.e. x + 2y = 5 — notice the -(-y) became +2y. A wrong sign on the 7 - 2 is just as fatal as one on the left.

Each equation is a straight line, and the solution is the point where the two lines cross. Elimination is just algebra's way of finding that crossing point without drawing anything. But two lines don't always cross once:

So the algebra quietly tells you the geometry: a normal answer means one crossing, a nonsense statement means the lines miss, and a trivially-true statement means they lie on top of each other.

Elimination doesn't stop at two equations. Scale it up to hundreds or thousands of unknowns and it becomes Gaussian elimination — the workhorse algorithm computers use to solve gigantic systems. Every time an engineer models the stresses in a bridge, an economist balances a market of many goods, or a graphics card works out how light bounces around a 3-D scene, somewhere underneath it is grinding through exactly this: multiply a row, add it to another, cancel a variable, repeat. You'll meet the grown-up version later when you learn to substitute and, further on, to handle whole grids of numbers at once. The idea you just learned scales all the way up.

See it explained

Sal Khan works through solving a system by adding the equations to eliminate a variable.