How to Solve by Substitution: Steps and Worked Examples

Back to the store window. Sign A says a notebook and a pencil together cost 7 dollars, so $x + y = 7$. Sign B says the notebook costs exactly one dollar more than the pencil, so $x = y + 1$. Sign B never mentions a price by itself - yet it hands you $x$ on a plate. Swap $y + 1$ in for $x$ inside Sign A and only one unknown is left: $(y + 1) + y = 7$, so $y = 3$ and $x = 4$. That swap is the entire solve by substitution method. When you solve by substitution, one equation donates an expression.

The two signs describe the same $x$. Sign B says $x = y + 1$, and Sign A agrees. Replacing $x$ with $y + 1$ inside Sign A swaps an equal quantity for an equal quantity. The truth of the equation never wobbles. This is exactly what it means to solve by substitution: use one equation to express one unknown in terms of the other. The swap has a name, solve by substitution, and it is legal for the plainest of reasons: equals may replace equals. Solve by substitution also earns its keep where graphing is clumsy. Fractional answers and far-off intersection points are no trouble for a swap.

You can solve the system by substitution in four short moves:

1. Pick a variable with coefficient 1 and solve its equation for that variable, so you get something like $y = 3x - 4$.
2. Swap the expression in. Replace that variable inside the other equation.
3. Solve the one-variable equation that remains.
4. Back-substitute. Carry the value into one original equation to get the second unknown, then check the pair in both originals.

If an equation already reads $y = \dots$ or $x = \dots$, step 1 is finished before you start. Every time you solve by substitution, the moves run in this order. Step 2 of solve by substitution is where beginners drift: the expression must land in the other equation, never back into its own. The first example below shows how easy solve by substitution gets when a variable arrives pre-solved.

To solve by substitution well, start with a smart pick. Any choice of variable works - the answer never depends on it. Some choices are just cheaper, so pick with intent. A variable wearing coefficient 1 unpacks with zero fractions. From $x - 2y = 0$ you read $x = 2y$ in one beat. A variable with coefficient $-3$ forces you to divide by $-3$ and drag fractions through every later step. Also look for a loose variable - one already sitting nearly alone, like the $+y$ in $4x + y = 24$. Moving it costs a single subtraction. In short: scan both equations first and let the friendliest variable go first. Solve by substitution that way, and you will never meet a single fraction.

Solve: $x + y = 12$ and $y = x + 2$.

Answer: $x = 5$, $y = 7$, so $(x, y) = (5, 7)$.

Steps. The second equation is already solved for $y$, so step 1 costs nothing. Swap $x + 2$ in for $y$ inside the first equation:

$x + (x + 2) = 12 \quad\Rightarrow\quad 2x + 2 = 12 \quad\Rightarrow\quad 2x = 10 \quad\Rightarrow\quad x = 5$

Back-substitute into the ready-made expression: $y = 5 + 2 = 7$. Check both originals: $5 + 7 = 12$ and $7 = 5 + 2$. Both hold. ✓

Notice what made this cheap: one equation arrived pre-solved, so solve by substitution landed immediately. When you see $y = \dots$ on the page, skip straight to the swap, and solve by substitution in three written lines.

Solve: $x - 2y = 0$ and $3x + 2y = 16$.

Answer: $x = 4$, $y = 2$, so $(x, y) = (4, 2)$.

Steps. Neither equation is ready for a swap yet, but to solve by substitution you need only one move: $x$ has coefficient 1 in the first. Rearrange it: $x = 2y$. Now swap $2y$ in for $x$ inside the second equation:

$3(2y) + 2y = 16 \quad\Rightarrow\quad 6y + 2y = 16 \quad\Rightarrow\quad 8y = 16 \quad\Rightarrow\quad y = 2$

Back-substitute into $x = 2y$: $x = 4$. Check the first original: $4 - 2(2) = 0$. ✓ And the second: $3(4) + 2(2) = 16$. ✓

The only new skill here was step 1 - rearranging before the swap. Solve by substitution this way and the algebra stays light. Want to check a system of your own? The [/calculators/substitution-calculator] performs the swap for any pair you type. Solve by substitution on paper first; then use the calculator as a second pair of eyes.

Solve: $3x - 2y = -2$ and $x + y = 6$.

Answer: $x = 2$, $y = 4$, so $(x, y) = (2, 4)$.

Steps. To solve by substitution here, notice the second equation carries two coefficient-1 variables. Solve it for $y$, the loose one: $y = 6 - x$. Swap into the first equation and mind the sign on the swap:

$3x - 2(6 - x) = -2 \quad\Rightarrow\quad 3x - 12 + 2x = -2 \quad\Rightarrow\quad 5x = 10 \quad\Rightarrow\quad x = 2$

Back-substitute: $y = 6 - 2 = 4$. Check: $3(2) - 2(4) = -2$ ✓ and $2 + 4 = 6$ ✓.

The distribution $-2(6 - x) = -12 + 2x$ is where most points leak, not the swapping itself. When both steps stay honest, you solve by substitution in four lines. That is the promise of solve by substitution: a full solution, right before anyone grades it.

Both methods crack the same systems, so the choice is about fit. Solve by substitution when one equation is already solved for a variable, or becomes so with one small move. Solve by elimination when both equations sit in standard form with matching or opposing coefficients. One add or subtract clears a variable, as the elimination page shows. There is overlap: you can solve by substitution even when coefficients argue. You can also rearrange into $y = \dots$ and then eliminate. The overview page how to solve a system of equations places both methods inside the bigger toolbox, next to graphing. Read it, and the choice to solve by substitution becomes a choice, not a default.

• Swapping back into the same equation. If you built $y = 6 - x$ from $x + y = 6$ and then swap into that same equation, you get $x + (6 - x) = 6$, which melts to $6 = 6$ - true, and totally useless. Solve by substitution only across two different equations.
• Skipping the distribution. In $3x - 2(6 - x)$, the $-2$ must hit both terms: $-12 + 2x$, not $-12 - 2x$ and not $-12 - x$.
• Sign slips while rearranging. From $x - 2y = 0$ the expression is $x = 2y$. Writing $x = -2y$ poisons every later step of solve by substitution.
• Quitting after one check. A pair is a solution only when both original equations hold. Checking one can hide an error in the other.
• Choosing a coefficient-3 variable when a coefficient-1 one is on offer. Nothing is wrong with fractions, but when you solve by substitution with a free choice available, take the fraction-free path.