To solve a system of two linear equations, find the one pair (x, y) that makes both equations true. With substitution, you solve one equation for a variable and plug that expression into the other. With elimination, you scale the equations so one variable has matching or opposite coefficients, then add or subtract to cancel it. Either way you end with one equation in one variable, solve it, and back-substitute for the other.
This guide works both methods step by step, solves one system both ways to show they are really the same calculation, and covers the two special cases (no solution and infinitely many) plus word problems with tickets and mixtures.
What Solving a System of Equations Means
(x, y) that satisfies every equation at the same time. On a graph, it is the point where the lines cross.Each linear equation in two variables is a straight line, and every point on that line makes the equation true. A point that lies on both lines makes both equations true, so it is the solution. That picture explains why there are only three possible outcomes: two lines cross once, never (they're parallel), or everywhere (they're the same line). OpenStax Intermediate Algebra names these an independent system, an inconsistent system and a dependent system (whose lines are coincident). Any system with at least one solution is called consistent.
Graphing shows the answer, but reading an exact crossing point like (2.75, −1.4) off a sketch is guesswork. Substitution and elimination give the exact values with algebra. If you want a refresher on drawing the lines themselves, see graphing linear equations.
The Substitution Method Step by Step
- Isolate one variable
Pick the variable that is easiest to get alone, ideally one with a coefficient of 1 or −1, so you avoid fractions.
- Substitute into the other equation
Replace that variable with the expression, in brackets. You now have one equation with one unknown.
- Solve for the remaining variable
Expand the brackets, collect like terms and solve.
- Back-substitute
Put the value into the isolated expression from step 1 to get the other variable.
- Check in both original equations
Not just the one you rearranged. More on why below.
Why is it legal? If y = 2x − 1 is true at the solution, then y and 2x − 1 are the same number there, so swapping one for the other in the second equation changes nothing at that point.
Solve y = 2x − 1 and 3x + 2y = 12.
The first equation already gives y. Substitute it into the second: 3x + 2(2x − 1) = 12.
Expand: 3x + 4x − 2 = 12, so 7x = 14 and x = 2.
Back-substitute: y = 2(2) − 1 = 3.
Check: 3 = 2(2) − 1 ✓ and 3(2) + 2(3) = 6 + 6 = 12 ✓.
(x, y) = (2, 3)The Elimination Method Step by Step
Elimination (also called the addition method) rests on one fact: if A = B and C = D, then A + C = B + D. Adding two true equations gives another true equation. Choose the multipliers well and one variable drops out of the sum.
- Line up the variables
Write both equations as
ax + by = c, with x terms, y terms and constants in columns. - Match one variable's coefficients
Multiply one or both equations (every term, including the constant) so the x or y coefficients are equal or opposite. The least common multiple keeps the numbers small.
- Add or subtract
Opposite coefficients: add. Equal coefficients: subtract. The chosen variable cancels.
- Solve, then back-substitute
Solve the one-variable equation and put the value into either original equation.
Solve 3x + 2y = 16 and 5x − 4y = 12.
The y coefficients are 2 and −4. Multiply the first equation by 2: 6x + 4y = 32. Now the y terms are opposites.
Add: (6x + 4y) + (5x − 4y) = 32 + 12, so 11x = 44 and x = 4.
Back-substitute into the first original equation: 3(4) + 2y = 16, so 2y = 4 and y = 2.
Check both: 3(4) + 2(2) = 12 + 4 = 16 ✓ and 5(4) − 4(2) = 20 − 8 = 12 ✓.
(x, y) = (4, 2)Solve 4x + 3y = 10 and 3x − 5y = 22.
No coefficient divides another, so target the x terms. The least common multiple of 4 and 3 is 12. Multiply the first equation by 3 and the second by 4: 12x + 9y = 30 and 12x − 20y = 88.
The x coefficients are equal, so subtract the second from the first. Every term of the second changes sign: 9y − (−20y) = 29y and 30 − 88 = −58. So 29y = −58 and y = −2.
Back-substitute: 4x + 3(−2) = 10, so 4x = 16 and x = 4.
Check both: 4(4) + 3(−2) = 16 − 6 = 10 ✓ and 3(4) − 5(−2) = 12 + 10 = 22 ✓.
(x, y) = (4, −2)When subtracting equations, students cancel the matched variable correctly and then subtract only the first term after it. In Example 3, writing 9y − 20y = −11y instead of 9y − (−20y) = 29y gives y = 58/11, a messy answer that fails the check. The fix: put the whole second equation in brackets before subtracting, or multiply it by −1 and add instead. A second classic slip is multiplying only the left side of an equation when scaling it; the constant on the right must be multiplied too.
One System Solved Both Ways
The two methods are usually taught as separate techniques, but they lead to the same equation. Take 2x + y = 7 and x − 3y = −7.
| Step | Substitution | Elimination |
|---|---|---|
| Prepare | From the first equation, y = 7 − 2x | Multiply the first equation by 3: 6x + 3y = 21 |
| Combine | x − 3(7 − 2x) = −7, so x − 21 + 6x = −7 | Add to x − 3y = −7: 7x + 0y = 14 |
| One-variable equation | 7x = 14 | 7x = 14 |
| Result | x = 2, y = 7 − 4 = 3 | x = 2, y = 7 − 4 = 3 |
Both columns reach 7x = 14. Substituting y = 7 − 2x into −3y does exactly what multiplying the first equation by 3 and adding does: it removes the y term using a multiple of the first equation. So the choice of method is about which version keeps your arithmetic cleaner, not about which one is "correct".
Substitution or Elimination: Which to Use
| What the system looks like | Use | Example |
|---|---|---|
| A variable is already alone | Substitution | y = 2x − 1, 3x + 2y = 12 |
| A variable has coefficient 1 or −1 | Substitution (no fractions appear) | 2x + y = 7, x − 3y = −7 |
Both in ax + by = c form, coefficients like 3 and 5 | Elimination | 4x + 3y = 10, 3x − 5y = 22 |
| Matching or opposite coefficients already | Elimination, add or subtract at once | x + y = 10, x − y = 4 |
| Fractions or decimals in the equations | Clear them first (multiply through), then elimination | Example 6 below |
A quick rule: if isolating a variable would force you to divide by something other than 1, elimination is usually the cleaner road. Standard form ax + by = c is the natural setup for elimination, while slope-intercept form y = mx + b hands you substitution for free.
No Solution or Infinitely Many Solutions
Sometimes both variables cancel at once. What remains tells you which special case you have:
- A false statement such as
0 = 7: no solution. The lines are parallel. - A true statement such as
0 = 0: infinitely many solutions. Both equations describe the same line.
(a) Solve 2x − 3y = 6 and 4x − 6y = 5. (b) Solve x − 2y = 4 and 3x − 6y = 12.
(a) Multiply the first equation by 2: 4x − 6y = 12. Subtract the second: 0 = 12 − 5 = 7. That is false, so there is no solution. In slope-intercept form the lines are y = (2/3)x − 2 and y = (2/3)x − 5/6: same slope, different intercepts, parallel.
(b) Multiply the first equation by 3: 3x − 6y = 12. Subtract the second: 0 = 0. Always true, so every point on the line is a solution. Write the set as y = (x − 4)/2, or as points (x, (x − 4)/2).
(a) no solution; (b) infinitely many, all points with y = (x − 4)/2You can also spot these cases before solving. For a₁x + b₁y = c₁ and a₂x + b₂y = c₂, cross-multiply the x and y coefficients. If a₁b₂ ≠ a₂b₁, the slopes differ and there is exactly one solution. If a₁b₂ = a₂b₁, the lines are parallel or identical, and you check whether one equation is an exact multiple of the other (constant included).
In 4(a), 2(−6) = −12 and 4(−3) = −12, so the lines are parallel, and since 2 × 6 = 12 ≠ 5, they never meet. The cross-multiplied form still works when a coefficient is 0, which the popular fraction version a₁/a₂ = b₁/b₂ does not.
Answer a special case in words, never as a made-up pair. "No solution" and "infinitely many solutions" are complete answers.
Systems of Equations Word Problems
Word problems become systems when they describe two unknowns with two facts. The routine: name each unknown with a letter and units, write one equation per fact, then solve. Usually one fact counts things (how many) and the other counts value (how much they cost, weigh or contain).
A school play sold 300 tickets. Adult tickets cost $12 and student tickets $7. Ticket sales came to $2,700. How many of each were sold?
Let a = adult tickets and s = student tickets. Count: a + s = 300. Money: 12a + 7s = 2700.
Substitution suits this one, since s = 300 − a. Then 12a + 7(300 − a) = 2700, so 12a + 2100 − 7a = 2700, 5a = 600 and a = 120.
Back-substitute: s = 300 − 120 = 180.
Check: 120 + 180 = 300 ✓ and 12(120) + 7(180) = 1440 + 1260 = 2700 ✓.
A lab has a 30% acid solution and an 80% acid solution. How many millilitres of each make 50 mL of a 60% solution?
Let x = mL of 30% solution and y = mL of 80% solution. Volume: x + y = 50. Pure acid: 0.30x + 0.80y = 0.60(50) = 30.
Clear the decimals by multiplying the acid equation by 10: 3x + 8y = 300. Multiply the volume equation by 3: 3x + 3y = 150.
Subtract: 5y = 150, so y = 30 mL. Then x = 50 − 30 = 20 mL.
Check: acid is 0.30(20) + 0.80(30) = 6 + 24 = 30 mL, and 30/50 = 60% ✓. Sanity check: 60% is closer to 80% than to 30%, so more of the 80% solution makes sense.
For a full routine on reading and translating word problems, see how to solve math word problems.
How to Check Your Answer
Substitute your pair into both original equations, the ones printed in the question. Checking only the equation you used for back-substitution proves almost nothing, because you built y to fit that equation; an error made earlier in the elimination step survives that check unnoticed.
A solution must satisfy every equation. If (4, −2) makes both original equations true, it lies on both lines, and two lines with different slopes cross at only one point. So a pair that passes both checks is the answer, however you found it.
Two more checks are worth a few seconds. First, a graph: plot both lines and see whether they cross near your point (on the Digital SAT, the Bluebook app has a built-in Desmos graphing calculator under the College Board calculator policy; see our Desmos SAT guide). Second, in word problems, ask whether the answer is sensible: a negative number of tickets or a mixture volume larger than the total means a setup error.
If your working and the check disagree and you can't find the slip, Solver AI on iPhone or iPad can scan your system and show its steps, so you can see the line where your work went a different way. Use it after you've tried the problem yourself, as a way to check, not to replace the practice.
Once two variables feel routine, the same elimination idea extends to three equations; see systems of equations with 3 variables. For bigger systems, the bookkeeping moves into matrices.
Practice Problems
Pick the method before you start, and check each answer in both equations.
- Solve
y = x + 4and2x + y = 13. Answer:(3, 7) - Solve
5x + 2y = 4and3x − 2y = 12. Answer:(2, −3)(add the equations) - Solve
6x − 4y = 8and9x − 6y = 12. Answer: infinitely many solutions, all points withy = (3x − 4)/2(both equations are multiples of3x − 2y = 4) - Solve
7x + 3y = 1and5x + 4y = −3. Answer:(1, −2)(multiply by 4 and 3, then subtract to remove y) - Solve
2(x − 3) + y = 3y + 4and(x + 1)/3 = y. Tidy each equation first. Answer:(8, 3)(the first becomesx − y = 5, the secondx = 3y − 1; substitute to get2y − 1 = 5)
Substitution when a variable is alone or has coefficient 1; elimination when the equations sit in ax + by = c form. Both reduce two equations to one. If everything cancels, 0 = 0 means infinitely many solutions and a false statement means none. Always check in both original equations.