Unit 4 · Lesson 2 of 4
Solving systems by elimination
You'll be able to: solve a system by elimination, and spot when it has no solution or infinitely many.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first, then check.
Solve by substitution: \(y = x - 1\) and \(2x + y = 8\).
Replace \(y\) with \((x - 1)\): \(2x + (x - 1) = 8\), so \(3x - 1 = 8\), \(3x = 9\), and \(x = 3\).
Then \(y = 3 - 1 = 2\).
Answer: \((3, 2)\).
Simplify \(-2(3x - 4) + 5x\).
Distribute the \(-2\) to both terms: \(-6x + 8 + 5x\).
Combine like terms: \(-6x + 5x = -x\).
Answer: \(-x + 8\).
Step 2 of 5The idea
Substitution works best when a letter is already alone. When both equations look like \(Ax + By = C\), the elimination method is usually faster.
The trick: if two equations are both true, you can add them, left side to left side and right side to right side. If one letter has opposite coefficients, like \(+3y\) and \(-3y\), it cancels out.
Elimination, step by step:
- Line up the equations: \(x\) under \(x\), \(y\) under \(y\), numbers on the right.
- Pick a letter to cancel. Multiply one or both equations so its coefficients are opposites.
- Add the equations. That letter disappears. Solve for the other one.
- Put that value into either original equation to find the second letter.
- Check the pair in both original equations.
If both letters cancel: a true statement like \(0 = 0\) means infinitely many solutions (same line). A false statement like \(0 = 7\) means no solution (parallel lines).
A system with no solution is called inconsistent. One with infinitely many is called dependent. For example, \(x + y = 2\) and \(x + y = 5\) can't both be true: subtracting gives \(0 = 3\).
Why this shows up on the placement test: test systems are often written as \(Ax + By = C\), where elimination is the quickest exact method, and "no solution" or "infinitely many" are common answer choices.
Step 3 of 5Worked examples
Example 1: Solve \(2x + 3y = 12\) and \(x - y = 1\)
- Cancel \(y\). The coefficients are \(3\) and \(-1\). Multiply the second equation by 3: \(3x - 3y = 3\). Now the \(y\) terms are \(+3y\) and \(-3y\).
- Add: \((2x + 3y) + (3x - 3y) = 12 + 3\), so \(5x = 15\).
- Solve: \(x = 3\).
- Use \(x - y = 1\): \(3 - y = 1\), so \(y = 2\).
- Check: \(2(3) + 3(2) = 12\) and \(3 - 2 = 1\). The solution is \((3, 2)\).
Example 2: Solve \(3x + 2y = 4\) and \(2x + 5y = -1\)
- No coefficient is an easy match, so multiply both. To cancel \(x\): first equation times 2, second times \(-3\). \(2 \cdot 3x = 6x\) and \(-3 \cdot 2x = -6x\) are opposites.
- That gives \(6x + 4y = 8\) and \(-6x - 15y = 3\).
- Add: \(-11y = 11\), so \(y = -1\).
- Use \(3x + 2y = 4\): \(3x + 2(-1) = 4\), so \(3x = 6\) and \(x = 2\).
- Check: \(2(2) + 5(-1) = 4 - 5 = -1\). True. The solution is \((2, -1)\).
Step 4 of 5Practice
Solve each system by elimination. Write points like \((3, 2)\). Type none for no solution or infinite for infinitely many.
\(x + y = 8\) and \(x - y = 2\)
The \(y\) terms are already opposites. Add: \(2x = 10\), so \(x = 5\).
Then \(5 + y = 8\), so \(y = 3\).
Answer: \((5, 3)\).
\(3x + 2y = 7\) and \(-3x + 4y = 5\)
The \(x\) terms are opposites. Add: \(6y = 12\), so \(y = 2\).
Then \(3x + 2(2) = 7\), so \(3x = 3\) and \(x = 1\).
Answer: \((1, 2)\).
\(2x + y = 5\) and \(3x - 2y = 4\)
Multiply the first equation by 2: \(4x + 2y = 10\).
Add it to \(3x - 2y = 4\): \(7x = 14\), so \(x = 2\).
Then \(2(2) + y = 5\), so \(y = 1\).
Answer: \((2, 1)\).
\(x + 3y = -2\) and \(2x - 5y = 7\)
Multiply the first equation by \(-2\): \(-2x - 6y = 4\).
Add it to \(2x - 5y = 7\): \(-11y = 11\), so \(y = -1\).
Then \(x + 3(-1) = -2\), so \(x = 1\).
Answer: \((1, -1)\).
\(2x + 3y = 1\) and \(3x - 4y = -7\)
Cancel \(y\): first equation times 4, second times 3. That gives \(8x + 12y = 4\) and \(9x - 12y = -21\).
Add: \(17x = -17\), so \(x = -1\).
Then \(2(-1) + 3y = 1\), so \(3y = 3\) and \(y = 1\).
Answer: \((-1, 1)\).
\(x - 2y = 4\) and \(-2x + 4y = 3\)
Multiply the first equation by 2: \(2x - 4y = 8\).
Add it to \(-2x + 4y = 3\): \(0 = 11\). Both letters cancel and the statement is false.
Answer: none (the lines are parallel).
\(3x + y = 2\) and \(6x + 2y = 4\)
Multiply the first equation by \(-2\): \(-6x - 2y = -4\).
Add it to \(6x + 2y = 4\): \(0 = 0\). Both letters cancel and the statement is true.
Answer: infinitely many solutions (the second equation is just the first one times 2, so it is the same line).
Solve the system \(5x - 2y = 16\) and \(3y = 2x - 2\).
Line up the second equation first: subtract \(2x\) from both sides to get \(-2x + 3y = -2\).
Cancel \(y\): first equation times 3, second times 2. That gives \(15x - 6y = 48\) and \(-4x + 6y = -4\).
Add: \(11x = 44\), so \(x = 4\). Then \(5(4) - 2y = 16\), so \(-2y = -4\) and \(y = 2\).
Answer: \((4, 2)\).
Step 5 of 5Wrap-up
You can now solve a system by multiplying and adding equations so one letter cancels, and read \(0 = 0\) as infinitely many solutions and \(0 = 7\) as none.
Read or watch next: OpenStax Intermediate Algebra 2e, 4.1 Solve Systems of Linear Equations with Two Variables (the part "Solve a System of Equations by Elimination"; free textbook, with more practice at the end of the section).
Also good: Khan Academy: Solving systems of equations with elimination (short videos). Quick rules: Lines and systems cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "How do I choose which letter to cancel, and what to multiply by?" It knows this course and can explain things another way.