Guides / Systems of equations

Solving systems of equations: substitution vs elimination

Both methods always work. Picking the one that fits the problem is what keeps the arithmetic short, and short arithmetic is where fewer sign mistakes live.

What "solving a system" means

Two linear equations in x and y are two lines. Solving the system means finding the point where the lines cross: the one (x, y) pair that makes both equations true. Substitution and elimination are two routes to that same point. Neither is "more correct."

Example 1: substitution, because y is already isolated

Solve y = 2x − 1 and 3x + y = 14.

1. 3x + (2x − 1) = 14Replace y in the second equation with what the first equation says y equals. Keep the parentheses on the first line.
2. 5x − 1 = 14Combine like terms.
3. x = 3Add 1, divide by 5.
4. y = 2(3) − 1 = 5Put x back into the equation that was already solved for y. Solution: (3, 5).

Check in the equation you did not substitute back into: 3(3) + 5 = 14. True.

Example 2: elimination, because the coefficients already cancel

Solve 2x + 3y = 12 and 4x − 3y = 6.

The y terms are +3y and −3y. Adding the equations makes them disappear.

1. 6x = 18Add the two equations, left side to left side and right side to right side. 3y + (−3y) = 0.
2. x = 3Divide by 6.
3. 2(3) + 3y = 12Put x = 3 into either original equation.
4. y = 26 + 3y = 12, so 3y = 6. Solution: (3, 2).

Check in the other equation: 4(3) − 3(2) = 12 − 6 = 6. True.

Example 3: elimination with a multiplier

Solve 3x + 2y = 16 and 5x − 4y = −10.

Nothing cancels yet. But 2y times 2 is 4y, which would cancel −4y. So multiply the whole first equation by 2.

1. 6x + 4y = 32First equation × 2. Every term, including the 16.
2. 11x = 22Add to the second equation: 6x + 5x, 4y − 4y, 32 + (−10).
3. x = 2Divide by 11.
4. 3(2) + 2y = 16Back into the original first equation.
5. y = 56 + 2y = 16, so 2y = 10. Solution: (2, 5).

Check: 5(2) − 4(5) = 10 − 20 = −10. True.

Sometimes both equations need a multiplier. For 2x + 5y = 1 and 3x − 2y = 11, multiply the first by 3 and the second by 2 to get 6x in both, then subtract. Or multiply the first by 2 and the second by 5 to get 10y and −10y, then add. Adding is safer than subtracting (see mistake 2 below), so when you have the choice, aim for opposite signs and add.

Example 4: substitution when nothing is isolated yet

Solve x − 3y = 7 and 2x + y = 7.

Elimination works fine here. But x has coefficient 1 in the first equation, so isolating it costs one step and no fractions.

1. x = 3y + 7Add 3y to both sides of the first equation.
2. 2(3y + 7) + y = 7Substitute into the second equation. Parentheses again.
3. 6y + 14 + y = 7Distribute the 2 over both terms.
4. 7y = −7Combine, subtract 14.
5. y = −1Divide by 7.
6. x = 3(−1) + 7 = 4Solution: (4, −1).

If instead you had isolated y from the first equation, you would get y = (x − 7)/3 and carry a fraction through every line. Pick the variable whose coefficient is 1 or −1.

How to choose, in one table

• A variable is already alone on one side: substitution.
• A variable has coefficient 1 or −1: substitution, isolate that one.
• Both equations look like ax + by = c and nothing has coefficient 1: elimination.
• The same variable has opposite coefficients (3y and −3y): elimination, just add.
• Word problems: whatever you wrote down first usually decides it. "Total is 20" and "one is 4 more than the other" gives you x + y = 20 and x = y + 4, which is set up for substitution.

Example 5: when the lines don't cross

Solve 2x + y = 5 and 4x + 2y = 7.

1. −4x − 2y = −10First equation × (−2), to cancel the 4x.
2. 0 = −3Add to the second equation. Both variables vanish and what's left is false.

A false statement with no variables means no solution: the lines are parallel. If you had gotten 0 = 0 instead, a true statement with no variables, the two equations are the same line and there are infinitely many solutions. Either way, the variables disappearing is the answer, not a sign you made an error. Check by comparing slopes: both lines here have slope −2 and different intercepts.

Common mistakes, and the exact line they happen on

1. Dropping the parentheses when substituting

Substitute y = x + 3 into 2x − y = 1.

1. 2x − x + 3 = 1This is the mistake. The minus applies to all of y, so it applies to the 3 as well. It should be 2x − (x + 3) = 1, which is 2x − x − 3 = 1, so x = 4, not x = −2.

Write the substituted expression inside parentheses on the first line, every time, even when it turns out not to matter. It costs nothing when the sign in front is plus, and it saves you when it's minus.

2. Subtracting equations and only subtracting one side

Eliminate y from 5x + 2y = 20 and 3x + 2y = 16 by subtracting.

1. 2x = 36This is the mistake. The left sides were subtracted (5x − 3x) but the right sides were added (20 + 16). Subtract both: 2x = 4, so x = 2.

Subtraction also invites a second slip: 5x − 3x is easy, but 2y − 2y next to −4y − (−4y) is where signs get lost. The fix that removes both problems: multiply one equation by −1 and add. Adding never has a "which side am I subtracting from" question.

3. Multiplying only some of the terms

1. 3x + 2y = 16Want to multiply by 2.
2. 6x + 4y = 16This is the mistake. The right side has to be multiplied too: 6x + 4y = 32. This is the same forgotten-term error as when clearing fractions.

4. Stopping at one variable

Finding x = 3 is half the answer. The question asks where the lines cross, and that is a point. Substitute back, write the pair, and check it in the equation you didn't use to find the second variable. If you used equation 1 to get y, checking in equation 1 will always "work" and tells you nothing.

5. Substituting back into the wrong place

After finding x, substitute into one of the original equations, or into the isolated form you built from an original. Substituting into a line you made partway through the elimination, like 6x + 4y = 32, is fine mathematically, but it's the line most likely to already carry an arithmetic error.

Practice

1. Solve y = 3x − 4 and x + y = 8.

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   Substitution: x + (3x − 4) = 8, 4x = 12, x = 3, y = 5. Solution (3, 5).

2. Solve x + 4y = 9 and x − 4y = 1.

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   Elimination by adding: 2x = 10, x = 5, then 5 + 4y = 9, y = 1. Solution (5, 1).

3. Solve 2x + 3y = 7 and 4x − y = 7.

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   Multiply the second by 3: 12x − 3y = 21. Add: 14x = 28, x = 2. Then 4(2) − y = 7, y = 1. Solution (2, 1). (Or isolate y = 4x − 7 and substitute.)

4. Solve x − 2y = 3 and 3x + y = 2.

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   x = 2y + 3, so 3(2y + 3) + y = 2, 7y + 9 = 2, y = −1, x = 1. Solution (1, −1).

5. Solve 3x − 6y = 9 and x − 2y = 3.

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   Multiply the second by −3: −3x + 6y = −9. Add: 0 = 0. Same line, infinitely many solutions.

6. Solve 5x + 2y = 4 and 3x + 4y = −6.

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   Multiply the first by −2: −10x − 4y = −8. Add: −7x = −14, x = 2. Then 10 + 2y = 4, y = −3. Solution (2, −3).