How to Solve Systems of Equations (3 Methods)

Each linear equation in two variables is a line. The solution to the system is the point where the lines cross, because that point sits on both lines.

Take this system:

$$y = 2x + 1 \qquad 3x + y = 11$$

Rewrite the second equation as $y = -3x + 11$. If you graph both lines, they meet at $(2, 5)$. Check it: $2(2) + 1 = 5$ and $3(2) + 5 = 11$. Both are true.

Graphing is great for intuition, but it is hard to read exact answers like $x = \frac{7}{3}$ off a graph. That is why algebra methods are usually faster.

Substitution works best when one equation already has a variable by itself.

Example: Solve $y = 2x + 1$ and $3x + y = 11$.

1. The first equation already gives $y$. Substitute $2x + 1$ for $y$ in the second equation: $3x + (2x + 1) = 11$.
2. Combine like terms: $5x + 1 = 11$.
3. Solve: $5x = 10$, so $x = 2$.
4. Plug $x = 2$ back into $y = 2x + 1$: $y = 2(2) + 1 = 5$.
5. Check in the other equation: $3(2) + 5 = 11$. Correct.

The solution is $(2, 5)$, which matches the graph.

Elimination works best when both equations are in standard form, $Ax + By = C$. You add or subtract the equations so one variable cancels.

Example 1: Solve $2x + 3y = 12$ and $4x - 3y = 6$.

1. The $y$ terms are $+3y$ and $-3y$, so adding the equations cancels $y$.
2. Add: $(2x + 4x) + (3y - 3y) = 12 + 6$, which gives $6x = 18$.
3. Solve: $x = 3$.
4. Substitute into the first equation: $2(3) + 3y = 12$, so $3y = 6$ and $y = 2$.
5. Check: $4(3) - 3(2) = 12 - 6 = 6$. Correct.

The solution is $(3, 2)$.

Example 2: Solve $3x + 2y = 16$ and $2x - y = 6$.

1. Nothing cancels yet. Multiply the second equation by 2: $4x - 2y = 12$.
2. Add it to the first equation: $7x = 28$, so $x = 4$.
3. Substitute into $2x - y = 6$: $8 - y = 6$, so $y = 2$.
4. Check: $3(4) + 2(2) = 12 + 4 = 16$. Correct.

The solution is $(4, 2)$.

Not every system has exactly one answer. The algebra tells you which case you are in.

No solution: Solve $x + y = 4$ and $x + y = 7$. Subtracting gives $0 = 3$, which is false. The lines are parallel (same slope, different intercepts), so they never meet.

Infinitely many solutions: Solve $2x + y = 5$ and $4x + 2y = 10$. Multiply the first equation by 2 and you get the second equation exactly. Subtracting gives $0 = 0$, which is always true. Both equations describe the same line, so every point on it is a solution.

Rule of thumb:

• A false statement like $0 = 3$ means no solution.
• A true statement like $0 = 0$ means infinitely many solutions.

• Substitution: one variable is already isolated, like $y = 2x + 1$ or $x = 4 - y$.
• Elimination: both equations are in standard form, or the coefficients line up nicely.
• Graphing: you want to see what is happening, or the problem asks for a sketch.

All three methods give the same answer. Pick the one that means the least work, and always plug your answer into both original equations to check it.