You have solved systems by substitution. Elimination is a second method that is often faster when the equations are in standard form. The idea: combine the two equations so that one variable disappears.
Why it works
If $x + y = 7$ and $x - y = 1$ are both true, then adding the left sides gives the same number as adding the right sides: equal amounts added to equal amounts stay equal. So $(x + y) + (x - y) = 7 + 1$, which is $2x = 8$. Any solution of the system also solves this new equation, and the $y$ terms are gone. That is the move: add or subtract whole equations so that one variable cancels, solve for the other, then substitute back.
Key idea
If a variable has opposite coefficients ($3y$ and $-3y$), add the equations. If it has equal coefficients ($3y$ and $3y$), subtract them, term by term on both sides. Solve the one-variable equation, substitute back, and check in both equations.
Worked examples
Worked example: Opposites: add
Solve the system $$\begin{aligned} 3x + 2y &= 16 \\ 5x - 2y &= 0 \end{aligned}$$
- The $y$ terms, $2y$ and $-2y$, are opposites. Add: $8x = 16$, so $x = 2$.
- Substitute into the first: $3(2) + 2y = 16$, so $2y = 10$ and $y = 5$.
- Check in the second: $5(2) - 2(5) = 0$.
Answer: $(2, 5)$
Worked example: Equal: subtract
Solve the system $$\begin{aligned} 4x + 3y &= 10 \\ x + 3y &= -5 \end{aligned}$$
- The $y$ terms are both $3y$. Subtract the second equation from the first, every term: $(4x - x) + (3y - 3y) = 10 - (-5)$.
- $3x = 15$, so $x = 5$.
- Substitute: $5 + 3y = -5$, so $3y = -10$ and $y = -\frac{10}{3}$.
- Check in the first: $4(5) + 3\left(-\frac{10}{3}\right) = 20 - 10 = 10$.
Answer: $\left(5, -\frac{10}{3}\right)$
Common mistake
Subtracting one equation from the other but changing the sign of only the first term. Subtracting $x + 3y = -5$ means subtracting $x$, subtracting $3y$, and subtracting $-5$, which adds $5$ on the right. Writing it out with parentheses, $(4x + 3y) - (x + 3y) = 10 - (-5)$, keeps every sign right. Then check your answer in both original equations.