Solving Systems by Elimination

Elimination adds or subtracts equations so opposite coefficients cancel one variable.

Why adding equations works

If two equalities are true, their left sides and right sides can be added. For a system, combining equations creates another equation satisfied by every common solution.

Opposite terms such as $3y$ and $-3y$ sum to zero, revealing the remaining variable.

A reliable method

1. Write equations with like terms aligned.
2. Multiply one or both complete equations if needed to create opposite coefficients.
3. Add the equations.
4. Solve for the remaining variable.
5. Substitute back to find the other variable.
6. Check both original equations.

Multiplying an equation means multiplying every term on both sides.

Immediate cancellation

The opposite $y$-coefficients made no scaling necessary.

Multiply one equation

For $x+2y=7$ and $3x-y=7$, multiply the second equation by $2$ to get $6x-2y=14$. Adding to the first gives $7x=21$, so $x=3$. Then $3+2y=7$, giving $y=2$.

Choose small multipliers that create opposites with minimal arithmetic.

Multiply both equations

For coefficients $2x$ and $3x$, multiply equations by $3$ and $-2$ respectively, or use the least common multiple $6$. Keep an organized vertical layout so each coefficient remains attached to its term.

Subtracting one equation is equivalent to adding its negative; either notation is valid if signs are handled consistently.

Fractions and decimals

Elimination can clear awkward coefficients first. Multiplying an equation by a common denominator converts fractions to integers without changing its line. With decimals, powers of ten may simplify arithmetic.

No solution

If variables cancel and a false statement remains, such as $0=5$, the lines are parallel and no ordered pair satisfies both.

Infinitely many solutions

If elimination gives $0=0$, the equations are equivalent. Every point on their shared line is a solution. Confirm that all coefficients and constants are proportional.

Context example

Suppose $2$ adult tickets and $3$ student tickets cost $55$, while $3$ adult and $2$ student tickets cost $65$. Equations $2a+3s=55$ and $3a+2s=65$ can be scaled to eliminate either price. Solving gives $a=17$ and $s=7$ dollars.

Common mistakes

Multiplying only one term. Scale the whole equation.

Adding coefficients but not constants. Combine every aligned column and both right sides.

Forgetting back-substitution. One variable is not a full solution.

Interpreting $0=0$ as $(0,0)$. It means infinitely many solutions.

Quick self-check

• Are like terms aligned?
• Which coefficients can become opposites most efficiently?
• Did scaling affect every term?
• Have I found and checked both coordinates?

Related topics

• Systems of Linear Equations
• Solving Systems by Substitution
• Linear Equation
• Standard Form of a Line