The Substitution Method
A rectangular vegetable patch is twice as long as it is wide, and 30 m of fencing runs right
round it — how wide is it? Because the length is already spelled out in terms of the width, you
can slide that straight into the fencing equation and solve for one unknown at a time. That
slide is the whole idea of the substitution method.
Two equations, two unknowns. The
elimination method
adds or subtracts the equations to cancel a variable. Substitution takes a
different route: get one variable on its own in one equation, then put that expression in
place of the variable in the other. Two unknowns become one.
It shines when an equation already hands you a variable for free. Suppose a shop tells you
"a pen costs one pound less than twice a pencil" — that's y = 2x - 1
already — and separately that "three pencils and a pen cost nine pounds", i.e.
3x + y = 9. You know exactly what y is worth
in terms of x, so why carry two letters around? Swap it in.
y = 2x - 1 \qquad 3x + y = 9
The first equation already has y by itself, so it tells us exactly
what y is worth:
substitute
2x - 1 for y in the second equation and
only x is left.
The recipe is always the same:
-
Isolate one variable in one equation — make it the subject (this is just
rearranging the formula).
- Substitute that expression into the other equation.
- Solve the single-variable equation you now have.
- Back-substitute that value to find the second variable.
Pick whichever variable is already alone — or easiest to isolate — so you avoid fractions.
Here y is already the subject, so substitution is the natural choice.
Doing it: 3x + (2x - 1) = 9, so 5x - 1 = 9,
giving 5x = 10 and x = 2. Then back into the
first equation: y = 2(2) - 1 = 3. Solution
x = 2,\ y = 3.
See it solved
Step through the four moves on our pair — isolate, substitute, solve, back-substitute.
When you have to rearrange first
Often no variable is sitting alone yet. That's fine — spend one line making one the subject, then
carry on. Take:
x + 2y = 11 \qquad 3x - y = 5
Nothing is isolated, but the -y in the second equation is easy to free:
rearrange to y = 3x - 5. Now substitute that whole expression into the
first equation:
x + 2(3x - 5) = 11 \;\Rightarrow\; x + 6x - 10 = 11 \;\Rightarrow\; 7x = 21 \;\Rightarrow\; x = 3.
Back-substitute: y = 3(3) - 5 = 4. Solution
x = 3,\ y = 4, and the check holds:
3(3) - 4 = 5. Notice the brackets around
(3x - 5) — they made the 2 multiply
both terms.
A word problem: how old are they?
Word problems turn a story into two equations. "Amara is three times as old as her brother
Ben. In four years' time, their ages will add up to 32." Let
a be Amara's age now and b Ben's. The first
sentence hands you a variable already alone:
a = 3b \qquad (a + 4) + (b + 4) = 32
Substitution is the obvious tool — a is the subject of the first
equation, so drop 3b into the second:
(3b + 4) + (b + 4) = 32 \;\Rightarrow\; 4b + 8 = 32 \;\Rightarrow\; 4b = 24 \;\Rightarrow\; b = 6.
Ben is 6, so a = 3(6) = 18: Amara is
18. Check the story — in four years they'll be
22 and 10, adding to
32. Once the sentences become equations, it's the same four moves every
time.
A curve meets a line — substitution's home turf
Substitution really earns its keep when one equation isn't a straight line. Where does the line
y = x + 1 cross the curve y = x^2 - 1? Both
already give y on its own, so set the two expressions equal — that's
substitution in its purest form:
x^2 - 1 = x + 1 \;\Rightarrow\; x^2 - x - 2 = 0 \;\Rightarrow\; (x - 2)(x + 1) = 0.
The single equation became a
quadratic, so
there are two answers: x = 2 and
x = -1. Back-substitute each into y = x + 1:
the line meets the curve at (2, 3) and (-1, 0).
A line can cut a curve twice, so a quadratic's two roots are the two crossing points — exactly what
you'd expect from the picture.
-
Substitute into the OTHER equation. If you rearrange the first equation and
then put the result back into the same equation, everything collapses to
0 = 0 — true, but useless. It tells you nothing because you only ever
used one fact. Always feed the expression into the equation you haven't touched.
-
Use brackets. Substituting y = 2x - 1 into
3y means 3(2x - 1) = 6x - 3, not
6x - 1. Wrap the expression in brackets so every sign and
multiplication lands on the whole thing.
-
Find BOTH variables. Solving for x is only half the
answer. A solution is a pair — remember to back-substitute for the other letter.
The "line meets a curve" trick isn't just geometry homework. Fire a ball and its height follows a
curve; ask where it crosses the height of a target and you're solving a curve-meets-line system by
substitution. In economics, the demand curve and the supply curve are two
equations in price and quantity — substitute one into the other and the solution is the market
price where buyers and sellers agree. Anywhere two relationships have to hold at once, substitution
turns "where do they meet?" into a single equation you can crack.
Both methods always reach the same answer; picking the slicker one is a genuine skill. A quick
rule of thumb:
- If a variable is already alone (or has a coefficient of
1, so it's trivial to isolate), substitution is
usually fastest.
- If the coefficients line up neatly for adding or subtracting (like matching
2a terms), reach for
elimination.
- If one equation is a curve, substitution is nearly always the way in.
They're two routes up the same hill — learn both and you'll always spot the gentler slope.
See it explained
Sal Khan works a system by substituting one equation into the other.