Elimination
The elimination method solves simultaneous equations by adding or subtracting them so that one pronumeral disappears. Often one or both equations must first be multiplied so the coefficients match, as in \(3x+2y=7\) and \(5x-4y=8\). In NSW Year 10 Advanced maths the equations can include fractions, decimals, brackets and letters as coefficients (NESA outcome MA5‑EQU‑P‑02).
You will multiply one or both equations to match coefficients, rearrange equations into the form \(ax+by=c\), clear fractions and decimals, and use a quicker route by adding and subtracting a symmetric pair. You will find the coefficients of a line and of a parabola through two points, solve a digit problem, substitute \(a=\dfrac{1}{x}\) and \(b=\dfrac{1}{y}\), and find when a pair has infinitely many solutions. It is part of Year 10 Advanced Simultaneous Equations.
Every question with a fully worked solution.
- Elimination - Video - Solving simultaneous equations by elimination Watch
Theory
In the elimination method you add or subtract the two equations so that one pronumeral disappears. If the coefficients do not match, multiply one or both equations first. This Year 10 Advanced maths page covers matching coefficients with the LCM, clearing fractions and decimals, equations with \(\dfrac{1}{x}\) and \(\dfrac{1}{y}\), and pairs with no solution or infinitely many.
Elimination removes one pronumeral by combining the two equations. Adding equal amounts to both sides keeps an equation true, so you may add (or subtract) the left sides and the right sides of two equations.
The number in front of a pronumeral is its coefficient. If the coefficients of \(y\) are opposites, such as \(+5y\) and \(-5y\), add the equations. If they are equal, such as \(4x\) and \(4x\), subtract.
Multiplying every term of an equation by the same non-zero number gives an equivalent equation: the same line and the same solutions. Use it to make coefficients match.
Elimination suits equations written as \(ax + by = c\). Rearrange to that form first if needed.
To eliminate \(y\) from \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\), multiply so that both \(y\) coefficients become the LCM of \(b_1\) and \(b_2\):
The elimination method
- Tidy: expand brackets and move terms so each equation reads \(ax + by = c\). Clear fractions (multiply by the LCD) and decimals (multiply by \(10\) or \(100\)).
- Choose the pronumeral to eliminate: the one whose coefficients are already equal or opposite, or the one with the smaller LCM.
- Multiply one or both equations so its coefficients match. Multiply every term, including the right side.
- Add if the signs are opposite; subtract if they are the same. Solve the one-pronumeral equation.
- Substitute back into the simplest equation to find the other value, and check in an original equation.
Special cases and problems
- Both pronumerals vanish: \(0 = 0\) means infinitely many solutions (the same line); \(0 = 7\) means no solution (parallel lines).
- Reciprocals: replace \(\dfrac{1}{x}\) and \(\dfrac{1}{y}\) by \(a\) and \(b\), solve, then take reciprocals.
- Digit problems: a two-digit number with tens digit \(t\) and units digit \(u\) is \(10t + u\); reversed it is \(10u + t\).
- Exact answers: leave fractions such as \(x = \dfrac{17}{11}\) as fractions unless decimals are asked for.
No coefficients match. The \(y\) terms have opposite signs, so make them \(+12y\) and \(-12y\) (the LCM of \(4\) and \(3\)) and add.
| \((1) \times 3:\quad 9x + 12y\) | \(=\) | \(6 \quad (3)\) |
| \((2) \times 4:\quad 20x - 12y\) | \(=\) | \(52 \quad (4)\) |
| \((3) + (4):\quad 29x\) | \(=\) | \(58\) |
| \(x\) | \(=\) | \(2\) |
Substitute into (1):
| \(3(2) + 4y\) | \(=\) | \(2\) |
| \(4y\) | \(=\) | \(-4\) |
| \(y\) | \(=\) | \(-1\) |
Check in (2): \(5(2) - 3(-1) = 10 + 3 = 13\ \checkmark\)
| \(\therefore\ x\) | \(=\) | \(\boldsymbol{2},\quad y = \boldsymbol{-1}\) |
Clear the decimals and the fractions first, multiplying every term.
| \((1) \times 10:\quad 3x + 2y\) | \(=\) | \(30 \quad (3)\) |
| \((2) \times 12:\quad 3x - 4y\) | \(=\) | \(12 \quad (4)\) |
The \(x\) coefficients are equal, so subtract. Take care with \(2y - (-4y)\).
| \((3) - (4):\quad 6y\) | \(=\) | \(18\) |
| \(y\) | \(=\) | \(3\) |
| \(3x + 2(3)\) | \(=\) | \(30\) |
| \(x\) | \(=\) | \(8\) |
Check in the original (2): \(\dfrac{8}{4} - \dfrac{3}{3} = 2 - 1 = 1\ \checkmark\)
| \(\therefore\ x\) | \(=\) | \(\boldsymbol{8},\quad y = \boldsymbol{3}\) |
\(\dfrac{3}{x} = 3 \times \dfrac{1}{x} = 3a\), so the equations become linear in \(a\) and \(b\).
| \(3a + 4b\) | \(=\) | \(3 \quad (1)\) |
| \(6a - 2b\) | \(=\) | \(1 \quad (2)\) |
| \((2) \times 2:\quad 12a - 4b\) | \(=\) | \(2 \quad (3)\) |
| \((1) + (3):\quad 15a\) | \(=\) | \(5\) |
| \(a\) | \(=\) | \(\dfrac{1}{3}\) |
| \(3 \times \dfrac{1}{3} + 4b\) | \(=\) | \(3\) |
| \(b\) | \(=\) | \(\dfrac{1}{2}\) |
Now go back to \(x\) and \(y\). If \(\dfrac{1}{x} = \dfrac{1}{3}\) then \(x = 3\).
| \(x\) | \(=\) | \(\dfrac{1}{a} = \boldsymbol{3}\) |
| \(y\) | \(=\) | \(\dfrac{1}{b} = \boldsymbol{2}\) |
Check in (2): \(\dfrac{6}{3} - \dfrac{2}{2} = 2 - 1 = 1\ \checkmark\)
(i) Eliminate \(y\) to show that \((k + 2)x = m + 8\).
(ii) For which \(k\) and \(m\) are there infinitely many solutions?
(iii) For which are there none?
(iv) Solve the pair when \(k = 2\) and \(m = 4\), giving exact answers.
(i) Multiply (1) by \(2\) to get \(-6y\), then add (2).
| \((1) \times 2:\quad 2x - 6y\) | \(=\) | \(8\) |
| \((2) + \text{this}:\quad (k + 2)x\) | \(=\) | \(m + 8\) |
(ii) If \(k = -2\) the left side is \(0\). With \(m = -8\) the result is \(0 = 0\), true for every \(x\): the second equation is the first multiplied by \(-2\). Infinitely many solutions when \(\boldsymbol{k = -2,\ m = -8}\).
(iii) If \(k = -2\) but \(m \neq -8\), the result \(0 = m + 8\) is false: parallel lines, no solution when \(\boldsymbol{k = -2,\ m \neq -8}\).
(iv) With \(k = 2,\ m = 4\):
| \(4x\) | \(=\) | \(12\) |
| \(x\) | \(=\) | \(3\) |
| \(3 - 3y\) | \(=\) | \(4\) |
| \(y\) | \(=\) | \(-\dfrac{1}{3}\) |
| \(\therefore\ x\) | \(=\) | \(\boldsymbol{3},\quad y = \boldsymbol{-\dfrac{1}{3}}\) |
Check: \(2(3) + 6\left(-\dfrac{1}{3}\right) = 6 - 2 = 4\ \checkmark\)
Common pitfalls
Frequently asked questions
How do you solve simultaneous equations by elimination?
Make the coefficients of one pronumeral equal or opposite by multiplying the equations, then subtract or add the equations to remove it. Solve for the other pronumeral, substitute back, and check.
When do you add and when do you subtract?
Add when the coefficients have opposite signs, such as plus 3y and minus 3y. Subtract when they have the same sign, such as 4x and 4x.
Which pronumeral should I eliminate?
Choose the one whose coefficients already match or are opposites. Otherwise choose the one whose coefficients have the smaller lowest common multiple, so the numbers stay small.
Is elimination better than substitution?
Both give the same answer. Elimination is usually quicker when both equations are written as ax plus by equals c; substitution is quicker when one equation already reads y equals or x equals.
What if both pronumerals cancel out?
If you are left with a true statement such as 0 equals 0, the equations describe the same line and there are infinitely many solutions. If it is false, such as 0 equals 5, the lines are parallel and there is no solution.