Simultaneous Equations — Substitution & Elimination

For a pair of simultaneous equations, both substitution and elimination give the same answer — so the choice is purely about which one reaches a clean equation faster. Simultaneous equations are two or more equations solved together because the same values must satisfy all of them at once; in the usual case you solve two linear equations in two variables for one ordered pair $(x, y)$ that makes both true.

The two methods side by side

                  Substitution                    Elimination
Core move         isolate one variable,           add or subtract the equations
                  put it into the other           so one variable cancels
Fastest when      a variable is already           coefficients are already
                  by itself (e.g. y = 10 - x)     equal or opposites
Typical first     rearrange to make one           line up the equations and
step              variable the subject            combine them
Correctness       same ordered pair               same ordered pair

Neither method is more correct than the other. A solution is the point where both conditions hold — and for two lines, the point where they meet: cross once means one solution, parallel means none, same line means infinitely many.

When to use which

Use substitution when a variable is already isolated, or can be isolated without much rearranging. Use elimination when one variable cancels cleanly after adding or subtracting, especially when coefficients are already equal or opposites. The practical question is only which route gets you to a clean equation faster.

Worked example, solved both ways

Solve

$$x + y = 10$$

and

$$2x - y = 2$$

Elimination

The $y$ terms are opposites, so this system is friendly for elimination. Add the equations:

$$(x + y) + (2x - y) = 10 + 2$$ $$3x = 12 \quad\Rightarrow\quad x = 4$$

Substitute $x = 4$ into $x + y = 10$:

$$4 + y = 10 \quad\Rightarrow\quad y = 6$$

So $(x, y) = (4, 6)$.

Substitution

From the first equation, make $y$ the subject:

$$y = 10 - x$$

Substitute into the second:

$$2x - (10 - x) = 2$$ $$2x - 10 + x = 2$$ $$3x - 10 = 2 \quad\Rightarrow\quad 3x = 12 \quad\Rightarrow\quad x = 4$$

Then $y = 10 - 4 = 6$, giving $(x, y) = (4, 6)$ again. Both methods solve the same system, so they land on the same ordered pair.

Check in both equations

$$4 + 6 = 10 \qquad\text{and}\qquad 2(4) - 6 = 8 - 6 = 2$$

Both hold, so the solution is correct.

Common mistakes and confusion points

• Solving for only one variable. Finding $x$ is not enough if the question asks for the full pair.
• Losing a negative sign. In the example, $2x - (10 - x)$ must become $2x - 10 + x$, not $2x - 10 - x$.
• Choosing a method mechanically. If a variable is already isolated, substitution may be faster; if coefficients already cancel, elimination is cleaner. Picking the easier route reduces errors.
• Skipping the check. A wrong answer can still look tidy. Checking both equations is one of the fastest ways to catch a slip.

Apply it yourself

Solve

$$x + y = 13$$

and

$$3x - y = 7$$

first by elimination, then by substitution, and confirm both give the same ordered pair. After that, change the constants and notice which method becomes quicker. In school maths these show up in algebra, graphing, and word problems about totals, differences, prices, or mixtures, and the same idea extends to larger and non-linear systems.