How to Solve Simultaneous Equations: Substitution vs Elimination

Simultaneous equations are two (or more) equations that must be true at the same time. With two unknowns you need two equations, and the solution is the pair of values that satisfies both. This guide covers the two everyday methods — substitution and elimination — plus Cramer's rule, the special cases, and how to check your answer.

What "simultaneous" means

Each linear equation in $x$ and $y$ is a straight line. The solution is the point where the lines cross.

For $2x + 3y = 12$ and $x - y = 1$, the lines meet at $(3, 2)$: that is, $x = 3$ and $y = 2$. Graphs are a good picture but slow and imprecise; algebra gives exact answers.

Method 1 — Substitution

Solve one equation for one variable, then substitute that expression into the other equation.

Worked example: $2x + 3y = 12$ and $x - y = 1$

1. The second equation is easy to rearrange: $x = 1 + y$.
2. Substitute into the first: $2(1 + y) + 3y = 12$.
3. Expand and solve: $2 + 2y + 3y = 12 \Rightarrow 5y = 10 \Rightarrow y = 2$.
4. Back-substitute: $x = 1 + 2 = 3$.
5. Check both equations: $2(3) + 3(2) = 12$ ✓ and $3 - 2 = 1$ ✓.

Step 3 is just a one-variable equation — see solving one equation for x if that part is shaky.

Method 2 — Elimination

Add or subtract the equations so that one variable cancels out.

Worked example: $3x + 2y = 16$ and $5x - 2y = 16$

1. The $y$-coefficients are $+2$ and $-2$, so add the equations: $8x = 32$.
2. $x = 4$.
3. Substitute into the first equation: $3(4) + 2y = 16 \Rightarrow 2y = 4 \Rightarrow y = 2$.
4. Check: $5(4) - 2(2) = 20 - 4 = 16$ ✓.

When you must multiply first: $2x + 3y = 13$ and $3x + 2y = 12$

No coefficients match, so make them match. Multiply the first equation by 3 and the second by 2:

$$6x + 9y = 39 \qquad 6x + 4y = 24$$

Subtract: $5y = 15$, so $y = 3$. Then $2x + 9 = 13 \Rightarrow x = 2$. Check: $3(2) + 2(3) = 12$ ✓.

Which method should you use?

Both methods always give the same answer, so you can use one to check the other.

Method 3 — Cramer's rule (using determinants)

For $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:

$$x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix},\; D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix},\; D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}$$

For $2x + 3y = 12$, $x - y = 1$: $D = (2)(-1) - (3)(1) = -5$, $D_x = (12)(-1) - (3)(1) = -15$, $D_y = (2)(1) - (12)(1) = -10$. So $x = \frac{-15}{-5} = 3$ and $y = \frac{-10}{-5} = 2$ ✓. Cramer's rule needs $D \neq 0$ and extends to three unknowns using 3×3 determinants. The 2×2 determinant calculator speeds up the arithmetic.

No solution and infinitely many solutions

In both cases $D = 0$, which is why Cramer's rule can't be used.

A word problem

Two pens and three notebooks cost 12. A pen costs 1 more than a notebook. Find each price.

Let $x$ = pen price and $y$ = notebook price: $2x + 3y = 12$ and $x - y = 1$. That's our first example, so a pen costs 3 and a notebook costs 2. Check: $2(3) + 3(2) = 12$ ✓.

Common mistakes

• Substituting back into the equation you rearranged (it will always "check") — check in both original equations.
• Subtracting equations but forgetting to subtract every term, especially negatives.
• Multiplying only one side of an equation when scaling it.
• Stopping after finding one variable.

Use the simultaneous equations calculator

Enter both equations in the simultaneous equations calculator to see a full solution, or use the linear equation solver for the single-variable step.

FAQ

How do you solve simultaneous equations? Use substitution (rearrange one equation and substitute into the other) or elimination (add/subtract the equations to cancel one variable), then back-substitute and check.

Which is easier, substitution or elimination? Substitution when a variable has coefficient 1; elimination when coefficients match or can easily be made to match.

How do I know if there is no solution? If eliminating a variable leaves a false statement like $0 = 2$, the lines are parallel and there is no solution.

Can simultaneous equations have infinitely many solutions? Yes, when both equations describe the same line — eliminating gives $0 = 0$.

How do I check my answer? Substitute both values into both original equations.

Is there a simultaneous equations calculator with steps? Yes — the simultaneous equations calculator shows the full working.

Further reading: Wolfram MathWorld — Linear System of Equations · OpenStax Elementary Algebra 2e, chapter 5.