To solve a system of equations with fractions, clear the denominators in each equation, then solve the resulting system with whole numbers. In x/2 + y/3 = 4 and x/4 − y/6 = 0, the solution is x = 4, y = 6, after clearing each equation with its own LCM.
Each equation is cleared with the LCM of its own denominators, not with one number for the whole system. If you have a single equation with one unknown, start with the fraction equations calculator; here there are two unknowns and two equations. If you have an inequality, go to inequalities with fractions, where multiplying by a negative number flips the direction.
Which method to choose once the equations are cleared
Once the denominators are gone, the method depends on how the coefficients look. All of them reach the same pair.
| If, after clearing... | Method | First move and example |
|---|---|---|
| one unknown has equal or opposite coefficients | Elimination | Add or subtract the equations: x + y = 5 with x − y = 1 gives 2x = 6 |
| one unknown is isolated or equals 1 | Substitution | Plug the expression into the other equation: with y = 2x and x + y = 9, you get 3x = 9 |
| no unknown cancels | Elimination after multiplying | 2x + 3y = 7 and 3x − 2y = 4: multiply by 2 and by 3 to get 6y and −6y; adding gives 13x = 26 |
| the same unknown is easy to isolate in both | Setting equal | Solve for y in both and set them equal: with y = 2x − 1 and y = 11 − x, 2x − 1 = 11 − x |
The third row ends at x = 2, y = 1. If you are unsure, choose elimination: it avoids the intermediate fractions that appear when you isolate an unknown.
Worked example with two different LCMs
Worked example: x/2 + y/3 = 4 and x/4 − y/6 = 0
Find x and y so that both equations hold at the same time.
- Find the LCM of each equation. In the first, LCM(2, 3) = 6. In the second, LCM(4, 6) = 12.
- Clear the first. Multiply all three terms by 6: 6 · (x/2) + 6 · (y/3) = 6 · 4, which gives 3x + 2y = 24.
- Clear the second. Multiply by 12: 12 · (x/4) − 12 · (y/6) = 12 · 0, which gives 3x − 2y = 0.
- Add the two equations. The y terms cancel and you get 6x = 24, so x = 4.
- Solve for the other unknown. Substitute x = 4 into the second cleared equation: 12 − 2y = 0, so y = 6.
Result: x = 4, y = 6.
Check: in the starting equations, 4/2 + 6/3 = 2 + 2 = 4 and 4/4 − 6/6 = 1 − 1 = 0.
The first equation needed 6 and the second 12. Multiplying both by 12 would also work, but the first would become 6x + 4y = 48, with larger numbers and no advantage. To find the LCM of several denominators, you can use the least common multiple calculator.
When the solution is a fraction
Take x + y = 1 and x − y = 1/5. The second already has a single denominator, so multiplying by 5 is enough: 5x − 5y = 1. The first, times 5, is 5x + 5y = 5. Adding them gives 10x = 6 and x = 6/10, which simplifies to 3/5. Then y = 1 − 3/5 = 2/5.
Leave the solution as a fraction in lowest terms. Here 3/5 = 0.6 and 2/5 = 0.4 are terminating decimals, but with 1/3 a rounded decimal such as 0.33 no longer satisfies the equation (1/3 − 0.33 = 1/300); that is why you check with fractions: 3/5 + 2/5 = 1 and 3/5 − 2/5 = 1/5. Both original equations hold.
No solution or infinitely many solutions
Two lines can cross at one point, never cross, or be the same line. Clearing the equations shows which case you have.
| System | After clearing | Lines | Solutions |
|---|---|---|---|
| x/3 + y/3 = 1 and x + y = 4 | x + y = 3 and x + y = 4 | parallel | none (inconsistent) |
| x/5 + y/5 = 2 and x + y = 10 | x + y = 10 and x + y = 10 | the same line | infinitely many (dependent) |
| x/2 + y = 3 and x/2 − y = 1 | x + 2y = 6 and x − 2y = 2 | they cross | one: (4, 1) |
With elimination, an inconsistent system ends in a false statement such as 0 = 1, and a dependent one in an always-true statement such as 0 = 0. To tell in advance, compare the cleared equations: if the coefficients of x and y are proportional but the constant term does not follow that proportion, the system has no solution.
Why you can multiply by the LCM
Multiplying both sides of an equation by a nonzero number does not change its solutions. The LCM is just the number that turns every fraction into a whole number. Because it is positive, it also changes no signs, unlike what happens with an inequality.
On a graph, each equation is a line and the solution is the point where they cross. Clearing denominators rewrites the formula of the line with whole numbers, but the line does not move and the crossing point is the same. That is why the pair from the cleared equations also works in the original ones.
Decimals, parentheses and negative signs
A terminating decimal is converted to a fraction before clearing: 0.5x + y = 3 is x/2 + y = 3, and 0.25 is 1/4. Avoid rounding repeating decimals, such as 0.33 instead of 1/3, because that adds an error the fraction does not have.
If a numerator is a sum or a difference, keep the parentheses when multiplying. In (x − 1)/2 + y/3 = 2, multiplying by 6 gives 3(x − 1) + 2y = 12, which is 3x + 2y = 15. A negative coefficient can be removed by multiplying the whole equation by −1, which flips all its signs and leaves the solutions intact. If the equation mixes other operations, simplify it first with combined operations with fractions.
Practice with three systems
- Solve x/4 + y/2 = 3 and x − y = 0.
- Solve x/2 + y/4 = 2 and x/4 − y/2 = 0.
- Solve x/2 + y = 0 and x − y = 3. Watch the sign of y.
Show the answers
- Since x = y, the first becomes x + 2y = 12, that is 3x = 12: x = 4 and y = 4.
- You get 2x + y = 8 and x = 2y. Substituting, 5y = 8: y = 8/5 and x = 16/5.
- From the first, y = −x/2. Substituting into the second, 3x/2 = 3: x = 2 and y = −1. A negative result is valid; check that 2/2 + (−1) = 0 and 2 − (−1) = 3.
For practice with a single unknown, the missing fraction value tool solves equalities with an unknown term. For domain restrictions, see the guide to algebraic fractions and their domain.
Key points
Clear each equation with its own LCM, multiplying all the terms. Solve with elimination or substitution, leave the solutions as fractions in lowest terms, and check the pair in both original equations. A false statement at the end means an inconsistent system, and an always-true one means infinitely many solutions. More topics at this level are in the advanced content hub.
Sources
- GeoGebra, “Métodos de solución sistemas ecuaciones lineales 2 x 2” (in Spanish): presents the graphical, substitution, equalization, elimination and determinant methods for 2x2 systems (accessed October 17, 2026).
- Common Core State Standards Initiative, “Grade 8, Expressions and Equations,” standard 8.EE.C.8: solving pairs of simultaneous linear equations, with solutions as points of intersection (accessed October 17, 2026).