Solving by elimination — two equations, two unknowns, kill one letter first.
Solve for x and y
3x − 5y = 7
4x + 2y = 18
Scale to kill y: double the first and multiply the second by 5, giving 26x = 104, so x = 4. Then 16 + 2y = 18 makes y = 1. Check in the first: 12 − 5 = 7 ✓. B has the two values swapped.
Like terms must sit in the same column before anything else.
| Letter | Coefficients | Smallest match | Signs |
|---|---|---|---|
| x | 3 and 4 | 12 | same |
| y | 5 and 2 | 10 | opposite |
10 is smaller than 12, so the numbers stay light. The y signs are already opposite, so adding will wipe y out with no sign flipping.
10 is the smallest number that 5 and 2 both divide into.
So the multipliers are:
10 ÷ 5 = 2, so E1 is doubled. 10 ÷ 2 = 5, so E2 is multiplied by 5. Each equation gets its own multiplier.
Opposites — one minus, one plus.
The 7 and the 18 get multiplied too. Miss the right side and the line stops being the same equation.
6x − 10y = 14
+ 20x + 10y = 90
26x 0y = 104
The y column adds to zero.
26x = 104
the added line
x = 4
÷ 26 both sides
Use whichever original equation has the smaller numbers.
x becomes 4
4(4) + 2y = 18
swap x for 4
16 + 2y = 18
4 × 4 = 16
2y = 2
− 16 both sides
y = 1
÷ 2 both sides
If eliminating one letter wipes out the other too, the lines do not cross at one point. Compare the coefficients of a₁x + b₁y = c₁ and a₂x + b₂y = c₂.
| Coefficients | Lines | Solutions |
|---|---|---|
| a₁a₂ ≠ b₁b₂ | cross | exactly one |
| a₁a₂ = b₁b₂ ≠ c₁c₂ | parallel | none — you reach 0 = a non-zero number |
| a₁a₂ = b₁b₂ = c₁c₂ | the same line | infinitely many — you reach 0 = 0 |
For what k does 3x + ky = 4 and 6x − 8y = 1 have no solution?
6x is 2 × 3x, so 6x − 8y must be 2(3x + ky)
left sides in proportion
2k = −8 → k = −4
match the y terms
then E1 × 2: 6x − 8y = 8, but E2 says 6x − 8y = 1
8 ≠ 1: parallel, no solution ✓
Matching left sides is not enough on its own: if the constants matched as well (here, if E2 ended in = 8), the two equations would be the same line with infinitely many solutions — the opposite answer.
Solve 3y = 2x − 16 and 3x − 4y = 23.
2x − 3y = 16
E1 rearranged: x and y left, number right
6x − 9y = 48 and 6x − 8y = 46
× 3 and × 2: x terms match
(6x − 8y) − (6x − 9y) = 46 − 48 → y = −2
same signs: subtract
2x + 6 = 16 → (5, −2)
check: 3(−2) = 10 − 16 ✓
Different: the columns only exist after both equations are written as ax + by = c.
Rewriting 3y = 2x − 16 as 2x + 3y = 16 keeps the wrong sign on one term. Move each term across with its sign flipped: −2x + 3y = −16, or ×(−1), 2x − 3y = 16.
Solve x2 + y3 = 4 and x4 − y6 = 0.
3x + 2y = 24 and 3x − 2y = 0
× 6 and × 12, every term
6x = 24 → x = 4
add: y cancels
12 + 2y = 24 → (4, 6)
check: 2 + 2 = 4 ✓, 1 − 1 = 0 ✓
Different: clear the fractions in each equation first (each with its own LCD); then it is an ordinary elimination. A zero on the right is fine — it stays zero.
When the coefficients are swapped between the equations, adding gives x + y directly. If 3x + 5y = 21 and 5x + 3y = 19, find x + y.
8x + 8y = 40
add the two equations
x + y = 5
÷ 8 — no need for x and y
Subtracting instead gives −2x + 2y = 2, so y − x = 1. Together: x = 2, y = 3, and 6 + 15 = 21 ✓.
Letters in the denominator: treat 1x and 1y as the unknowns. Solve 4x + 5y = 2 and 8x − 5y = 1.
a = 1x, b = 1y: 4a + 5b = 2, 8a − 5b = 1
now an ordinary system
12a = 3 → a = 14; then 1 + 5b = 2 → b = 15
add: b cancels
x = 4, y = 5
flip back: x = 1a, y = 1b
Stopping at a = 14 and b = 15 answers for the reciprocals. Turn them back over — and check: 44 + 55 = 2 ✓, 84 − 55 = 1 ✓.
Elimination items include price word problems (“three apples and two mangoes”), “find x + y” (add or subtract the equations before solving anything), and “for what k is there no solution”. Read what is asked first — it is often not x and y.
One point fits both lines
that point is (4, 1)
Solve the system
4x + 3y = 27
3x + 2y = 19
Multiply the first by 2 and the second by 3 to match the y terms, then subtract: x = 3. Substituting gives 3y = 15, so y = 5. Confirm in the second equation: 9 + 10 = 19 ✓. Always test your pair in the equation you did not finish with.
Hint: both y terms come out positive this time, so subtract the two lines instead of adding them.
