Lesson · Systems and Matrices · Systems

Solving by Elimination

Adding one equation to another is legal, and if the multiples are chosen so that the coefficients of one variable are opposites, that variable leaves. What remains is one equation in one unknown, and the system you now have is the system you started with.

Open the interactive lesson Practice

3x + 4y = 10 3x + 4y = 10 5x − 2y = 8 × 2 → 10x − 4y = 16 --------------- add 13x = 26 → x = 2, then y = 1

One multiplication, one addition, one unknown gone

Core ideas

What this lesson establishes

Two operations are used, and both need a reason. The reason is the same in each case: the step can be undone.

Adding two equations adds equals to equals

If a = b and c = d, then a + c = b + d. Elimination is that one fact applied to two equations, and it is why the sum of two true equations is a true equation.

Multiplying an equation by a nonzero number changes nothing

The solution set survives, because dividing by the same number puts the equation back. Multiplying by 0 is forbidden: 0 = 0 is satisfied by every pair, and there is no operation that recovers what was there.

The new system is equivalent, not merely implied

Replacing the second equation by second plus k times first can be undone by subtracting k times the first. Nothing is lost and nothing is invented. Operations that cannot be undone — squaring both sides, in Rational and Radical Expressions — are exactly the ones that produce solutions that were never there.

The material

Elimination, and why the new system is equivalent

Two operations, the theorem that licenses them, and the three endings.

To eliminate a variable, multiply one or both equations by constants chosen so that the coefficients of that variable become opposites, then add the equations. The variable cancels, the sum is a new equation in the other variable alone, and solving it gives one coordinate.

Theorem · Elimination preserves the solution set

Let E₁ and E₂ be equations and let k be any number. The systems {E₁, E₂} and {E₁, E₂ + kE₁} have exactly the same solutions. If k ≠ 0, the systems {E₁, E₂} and {E₁, kE₂} have exactly the same solutions.

Suppose the pair (x, y) satisfies E₁ and E₂. Then the two sides of E₁ are equal numbers at that pair, so k times them are equal, and adding equal quantities to the two sides of E₂ leaves a true equation. So the pair satisfies E₂ + kE₁, and it still satisfies E₁, which was not touched.

Suppose instead the pair satisfies E₁ and E₂ + kE₁. Run the same argument with −k: adding −k times E₁ to E₂ + kE₁ gives E₂, so the pair satisfies E₂. Every solution of either system is a solution of the other.

For the second claim, multiplying E₂ by k is undone by multiplying by 1/k, which exists exactly when k ≠ 0.

□

Keeping E₁ unchanged is what makes the argument run in both directions. Replacing both equations at once by two combinations is not covered by this theorem, and it is genuinely unsafe: from x = 1 and y = 1 you may add each to the other and reach x + y = 2 twice, which has a whole line of solutions. One equation at a time, always.

Choosing the multipliers

Look at the coefficients of the variable you want to remove. If one divides the other, multiply the smaller equation and you are done. Otherwise take the least common multiple of the two coefficients: for 3y and −2y that is 6, so multiply by 2 and 3. The sign of one multiplier is chosen to make the two coefficients opposite.

2x + 3y = 5 × 2 → 4x + 6y = 10 5x − 2y = −16 × 3 → 15x − 6y = −48 ------------------- add 19x = −38 → x = −2 2(−2) + 3y = 5 → 3y = 9 → y = 3 check 5(−2) − 2(3) = −10 − 6 = −16 ✓ solution (−2, 3)

Either variable may be eliminated, and the answer does not depend on which. The work does. Before starting, look at all four coefficients and pick the variable that costs one multiplication rather than two.

When both variables disappear

The sum can lose both variables at once, leaving a statement about numbers. A false one means the system is inconsistent. A true one means the second equation carried no information the first did not already have, and the solution set is the whole line.

2x − 6y = 4 2x − 6y = 4 −x + 3y = 5 × 2 → −2x + 6y = 10 --------------- add 0 = 14 FALSE → no solution 2x − 6y = 4 2x − 6y = 4 −x + 3y = −2 × 2 → −2x + 6y = −4 --------------- add 0 = 0 TRUE → infinitely many

Example · Describing the dependent solution set

The second system above is x − 3y = 2 multiplied by 2 and by −1, so its solution set is that line: {(3t + 2, t)}. At t = 0 the pair is (2, 0) and at t = 1 it is (5, 1), and both satisfy 2x − 6y = 4 and −x + 3y = −2.

(0, 0) is not among them. "Infinitely many" is a count, and the answer is a set.

Between the two methods: elimination is quicker when both equations are in standard form and no coefficient is 1, which is most of the time. Substitution wins when a variable is already isolated or has a coefficient of 1. Neither is more correct, and the choice is worth five seconds before the pen moves.

Interactive

Solving by elimination

Multipliers from the lcm, and the unknown that cancels

Pick the multipliers

Choose which variable to eliminate and the lab shows the multiplied equations and their sum. Run the same system twice, once for each variable: two routes, one pair, and usually one of them is half the arithmetic.

Worked example

Equation 1

Equation 2

Which unknown to eliminate

——

Type equations the way you would write them: 2x + 3y = 12, y = 2x - 1, x/2 + y/3 = 1. Brackets and implicit multiplication are understood, and every coefficient becomes an exact fraction.

Method

Eliminating a variable

Four moves. The first is the one people skip and then pay for.

Line both equations up in standard form

ax + by = c, with the x term above the x term and the constants on the right. Elimination adds columns, and columns that are not aligned add the wrong things together.

Choose the variable and the two multipliers

Take the least common multiple of its two coefficients and give one equation the sign that makes them opposites. Choosing the cheaper variable here is the whole of the skill.

Add the equations and solve what is left

One variable, one equation, Linear Equations and Inequalities. If both variables vanish, read the numeric statement instead: false means no solution, true means infinitely many and the answer is a line.

Back-substitute into an original equation, then check in both

Use an equation you were given rather than one you multiplied — fewer chances to carry a stray factor. Then test the pair in both originals.

Elimination

Score 0 / 0

Choose the answer you can justify.

3x + 4y = 10 and 5x − 2y = 8

The y coefficients are 4 and −2. Doubling the second equation makes them 4 and −4, so one multiplication is enough and y goes on the first addition.

3x + 4y = 10 3x + 4y = 10 5x − 2y = 8 × 2 → 10x − 4y = 16 ---------------- add 13x = 26 x = 2 3(2) + 4y = 10 → 4y = 4 → y = 1 check 3(2) + 4(1) = 6 + 4 = 10 ✓ 5(2) − 2(1) = 10 − 2 = 8 ✓ solution (2, 1)

Eliminating x instead reaches the same pair by a longer road: multiply the first equation by 5 and the second by −3 to get 15x + 20y = 50 and −15x + 6y = −24, whose sum is 26y = 26. Two multiplications against one, with larger numbers in both. That difference is visible before any work is done, which is why the choice belongs at the start.

The 16 on the right of the doubled equation is the entry most often left behind. Multiplying an equation means multiplying the constant too, and a check in both originals catches the omission immediately.

Faded rehearsal: solve 4x + 3y = 1 and 2x − 3y = 11. The multipliers have already been chosen: both are 1, so add the aligned equations. Finish the remaining coordinate, check the pair in both originals, and say why eliminating x would cost more work.

Common mistakes

Adding equations that are not aligned3x + 4y = 10 written above 8 = 5x − 2y invites adding 4y to 5x. Standard form first, every time; the alignment is what makes the addition mean anything.

Multiplying only the left-hand side5x − 2y = 8 doubled is 10x − 4y = 16. The 16 is the part that gets left behind, and the resulting system is a different system that happens to look like the right one.

Reading 0 = 0 as x = 0It is a true statement with no variable in it, so the system is dependent and its solution set is a line. Reporting (0, 0) names a single pair which, for 2x − 6y = 4, is not even on that line.

Check your understanding

✓

Finish when you choose the cheaper variable before starting.

Look at the four coefficients, decide which variable costs one multiplication instead of two, and carry it through in exact fractions. Both routes give the same pair; being able to say which is shorter, and why, is the judgement that survives into “Gaussian Elimination”, where the choice is made for you by position on the page.

Note: Elimination is the rest of this half of the course. “Systems in Three Variables” performs it twice to remove one variable from a three-equation system; “Matrices and Row Operations” writes it without the letters, as operations on the rows of a matrix; “Gaussian Elimination” turns it into an algorithm with no decisions left in it.