Systems of equations tutorial
A system of equations is two or more equations that share the same variables. Solving the system means finding the values of $x$ and $y$ that satisfy both equations at the same time. This tutorial covers the two most common solution methods — substitution and elimination — and explains what to do in every case.
What you are looking for
Values of $x$ and $y$ such that both equations are true simultaneously — for example, the ordered pair $(x, y) = (2, 3)$ that satisfies both $x + y = 5$ and $x - y = -1$.
What is a system of equations?
Each equation in a system describes a straight line on the coordinate plane. The solution is the point (or points) where those lines cross. For a two-equation, two-variable system there are exactly three possibilities:
One solution — lines intersect
The lines cross at exactly one point. Most systems you will encounter in algebra have one solution. The ordered pair $(x, y)$ at the crossing point satisfies both equations.
No solution — lines are parallel
The lines have the same slope but different $y$-intercepts. They never cross. Writing the equations in slope-intercept form and seeing identical slopes is the fastest check.
Infinitely many solutions — same line
Both equations describe the exact same line. Every point on the line satisfies both equations simultaneously. This happens when one equation is a constant multiple of the other.
Three types of systems at a glance
The geometry of the lines determines how many solutions exist.
One solution
Lines intersect
No solution
Parallel lines
Infinite solutions
Same line
All practice questions on this site have exactly one integer solution. Learning to recognise the other two cases is still important for exams.
Method 1 — Substitution
Use substitution when one equation is already solved for a variable, or when it is easy to isolate one. The idea: replace a variable in one equation with an expression from the other.
- Choose the equation where a variable is isolated (or easiest to isolate).
- Write an expression for that variable: e.g. $y = 3x - 1$.
- Substitute that expression into the other equation, replacing every occurrence of $y$.
- Solve the resulting single-variable equation for $x$.
- Back-substitute the found $x$ value into either original equation and solve for $y$.
- Write the solution as an ordered pair $(x, y)$ and verify in both equations.
Method 2 — Elimination
Use elimination when both equations are in standard form $ax + by = c$. The idea: add (or subtract) the equations to cancel one variable entirely.
- Line up the equations so matching variables are in the same columns.
- Check whether any variable's coefficients are already opposites (e.g. $+3y$ and $-3y$).
- If not, multiply one or both equations by constants to create opposite coefficients.
- Add the two equations. One variable disappears.
- Solve the remaining single-variable equation.
- Substitute the found value back into either original equation and solve for the second variable.
The graphical picture
Every equation $ax + by = c$ is a line. The solution to a system is the ordered pair at the intersection point. The diagram below shows $x + y = 5$ (blue) and $x - y = -1$ (orange). They meet at $(2, 3)$.
Solving the system algebraically gives the same answer as reading the graph: $x = 2,\; y = 3$.
Worked examples
Example 1 — substitution
Solve: $\quad y = 2x - 1 \quad$ and $\quad 3x + y = 9$
Step 1. Equation 1 is already solved for $y$: $y = 2x - 1$.
Step 2. Substitute into Equation 2, replacing $y$:
$3x + (2x - 1) = 9$
$5x - 1 = 9$
$5x = 10$, so $x = 2$
Step 3. Back-substitute into Equation 1:
$y = 2(2) - 1 = 3$
Solution: $(x, y) = (2, 3)$
Check: $3(2) + 3 = 9$ ✓
Example 2 — elimination (direct)
Solve: $\quad 2x + 3y = 16 \quad$ and $\quad 4x - 3y = 2$
Step 1. The $y$-coefficients are $+3$ and $-3$ — already opposites. Add the equations:
$(2x + 3y) + (4x - 3y) = 16 + 2$
$6x = 18$, so $x = 3$
Step 2. Substitute $x = 3$ into the first equation:
$2(3) + 3y = 16 \;\Rightarrow\; 6 + 3y = 16 \;\Rightarrow\; 3y = 10 \;\Rightarrow\; y = \dfrac{10}{3}$
Not every system produces integers — this is a reminder to check your arithmetic if a fraction appears.
Check: $2(3) + 3\!\left(\tfrac{10}{3}\right) = 6 + 10 = 16$ ✓ $4(3) - 3\!\left(\tfrac{10}{3}\right) = 12 - 10 = 2$ ✓
Example 3 — elimination (multiply first)
Solve: $\quad x + 2y = 7 \quad$ and $\quad 3x - y = 0$
Step 1. Multiply Equation 2 by 2 to match the $y$-coefficient in Equation 1:
$6x - 2y = 0$
Step 2. Add to Equation 1 ($y$-terms cancel):
$(x + 2y) + (6x - 2y) = 7 + 0$
$7x = 7$, so $x = 1$
Step 3. Back-substitute into Equation 2:
$3(1) - y = 0 \;\Rightarrow\; y = 3$
Solution: $(x, y) = (1, 3)$
Check: $1 + 2(3) = 7$ ✓ $3(1) - 3 = 0$ ✓
Choosing the right method
Both substitution and elimination produce the same answer. Picking the faster route saves time, especially on timed tests.
| Situation | Faster method | Why |
|---|---|---|
| One equation is $y = \ldots$ or $x = \ldots$ | Substitution | Already isolated — plug straight in. |
| Coefficients of one variable are already opposites | Elimination | Add and the variable vanishes immediately. |
| Both equations in standard form $ax + by = c$ | Elimination | Multiply one row and add to cancel a variable. |
| Coefficients are messy fractions | Either — clear fractions first | Multiply each equation by its LCD to get integers before proceeding. |
Refreshable sample
Try one systems question
Work through the system before checking. Enter both $x$ and $y$, then click Check answer. The explanation shows the full algebraic steps whether you were right or wrong.
Common mistakes to avoid
- Substituting into the same equation you solved. After isolating $y$ in Equation 1, substitute into Equation 2 — not back into Equation 1. Substituting into the same equation produces a trivially true statement like $0 = 0$ and gives no new information.
- Forgetting to update the answer. After elimination gives you $x$, many students stop. You still need to back-substitute to find $y$. The solution requires both values.
- Sign errors when adding/subtracting equations. Each term carries its sign. Write out $+$ and $-$ signs explicitly — do not drop a negative when aligning columns.
- Multiplying only one side of an equation. When scaling an equation, every term on both sides must be multiplied. Multiplying only the left side changes the meaning of the equation.
- Not checking the answer. Substitute your $(x, y)$ pair back into both original equations. If either does not balance, an arithmetic error occurred somewhere.
- Assuming the system always has an integer solution. Real-world systems often produce fractions or decimals. In this context all practice questions have integer solutions, but that is not a property of systems of equations in general.