Solving Linear Systems by Elimination

When neither equation has a variable with a coefficient of $1$, substitution gets messy with fractions. Elimination avoids that. You add or subtract the two equations so that one variable cancels out, leaving a single equation in one variable. It’s often the fastest method when both equations are in the form $Ax + By = D$.

Key ideas

Why adding equations works

If $a = b$ and $c = d$, then $a + c = b + d$: adding equal amounts to both sides keeps an equation balanced. So you can add (or subtract) two equations, left side to left side and right side to right side, and the solution of the system still satisfies the result.

The trick is to choose the equations so that one variable cancels:

$$\begin{aligned} 3x + 2y &= 16 \\ 5x - 2y &= 0 \\ \hline 8x \phantom{{}+2y} &= 16 && \text{add: } 2y + (-2y) = 0 \end{aligned}$$

The method

1. Line up both equations in the form $Ax + By = D$, with like terms in columns.
2. Match a coefficient. If needed, multiply one or both equations by a number so that one variable has opposite coefficients (like $2y$ and $-2y$) or equal coefficients (like $3x$ and $3x$).
3. Eliminate. Add the equations if the coefficients are opposites; subtract if they’re equal.
4. Solve for the remaining variable.
5. Back-substitute into either original equation to find the other variable.
6. Check in both original equations.

Choosing which variable to eliminate

• If a variable already has opposite or equal coefficients, eliminate it.
• Otherwise, pick the variable whose coefficients are easiest to match. For $2x + 3y = 1$ and $4x - y = 9$, multiplying the second equation by $3$ turns $-y$ into $-3y$, the opposite of $3y$.
• When you must multiply both equations, aim for the lowest common multiple of the two coefficients. For $3x + 4y$ and $5x - 3y$, the LCM of $4$ and $3$ is $12$, so multiply by $3$ and $4$ to get $12y$ and $-12y$.
• Multiplying to get opposites lets you add, which avoids sign slips from subtracting.

No solution or infinitely many

Sometimes both variables disappear.

• A false statement, such as $0 = 18$: the system has no solution. The lines are parallel.
• A true statement, $0 = 0$: the system has infinitely many solutions. Both equations describe the same line, and every point on it is a solution.

These match the three cases you saw when graphing systems.

Substitution or elimination?

Use substitution when…	Use elimination when…
one equation is already solved for a variable, like $y = 2x - 7$	both equations are in the form $Ax + By = D$
a variable has coefficient $1$ or $-1$	no coefficient is $1$ or $-1$

Both methods always give the same answer, so pick whichever means less work.

Worked examples

Example 1: Coefficients that are already opposites

Solve the system.

$$\begin{aligned} 3x + 2y &= 16 \\ 5x - 2y &= 0 \end{aligned}$$

Solution. The $y$-terms are $2y$ and $-2y$, which are opposites. Add the equations:

$$8x = 16 \quad\Rightarrow\quad x = 2$$

Back-substitute into the first equation:

$$3(2) + 2y = 16 \quad\Rightarrow\quad 2y = 10 \quad\Rightarrow\quad y = 5$$

Check: $3(2) + 2(5) = 16$ ✓ and $5(2) - 2(5) = 0$ ✓. The solution is $(2, 5)$.

Example 2: Multiplying one equation

Solve the system.

$$\begin{aligned} 2x + 3y &= 1 \\ 4x - y &= 9 \end{aligned}$$

Solution. Multiply the second equation by $3$ so that its $y$-term becomes $-3y$:

$$\begin{aligned} 2x + 3y &= 1 \\ 12x - 3y &= 27 \end{aligned}$$

Add:

$$14x = 28 \quad\Rightarrow\quad x = 2$$

Back-substitute into the first equation:

$$2(2) + 3y = 1 \quad\Rightarrow\quad 3y = -3 \quad\Rightarrow\quad y = -1$$

Check: $2(2) + 3(-1) = 1$ ✓ and $4(2) - (-1) = 9$ ✓. The solution is $(2, -1)$.

Example 3: Multiplying both equations

Solve the system.

$$\begin{aligned} 3x + 4y &= 6 \\ 5x - 3y &= -19 \end{aligned}$$

Solution. Neither variable’s coefficients are multiples of each other, so multiply both equations. To eliminate $y$, make the $y$-terms $12y$ and $-12y$: multiply the first equation by $3$ and the second by $4$.

$$\begin{aligned} 9x + 12y &= 18 \\ 20x - 12y &= -76 \end{aligned}$$

Add:

$$29x = -58 \quad\Rightarrow\quad x = -2$$

Back-substitute into the first original equation:

$$3(-2) + 4y = 6 \quad\Rightarrow\quad 4y = 12 \quad\Rightarrow\quad y = 3$$

Check: $3(-2) + 4(3) = -6 + 12 = 6$ ✓ and $5(-2) - 3(3) = -10 - 9 = -19$ ✓. The solution is $(-2, 3)$.

(You could also eliminate $x$ by multiplying by $5$ and $3$ and subtracting. You’d get the same answer.)

Example 4: When both variables disappear

Solve each system.

• (a) $2x - 3y = 4$ and $-4x + 6y = 10$
• (b) $x + 2y = 3$ and $3x + 6y = 9$

Solution.

(a) Multiply the first equation by $2$ to get $4x - 6y = 8$, then add it to the second equation:

$$\begin{aligned} 4x - 6y &= 8 \\ -4x + 6y &= 10 \\ \hline 0 &= 18 \end{aligned}$$

That’s false, so the system has no solution. (Both lines have slope $\tfrac{2}{3}$ but different $y$-intercepts: they’re parallel.)

(b) Multiply the first equation by $3$ to get $3x + 6y = 9$. That’s exactly the second equation, so subtracting gives

$$0 = 0$$

That’s always true, so the system has infinitely many solutions: every point on the line $x + 2y = 3$, such as $(3, 0)$, $(1, 1)$ and $(-1, 2)$.

Common mistakes

Subtracting only the first term. When you subtract one equation from another, subtract every term, including the constant: $(3x - 4y) - (3x + 2y) = -6y$, not $-2y$. If subtraction feels risky, multiply one equation by $-1$ and add instead.

Multiplying only one side. Multiplying $4x - y = 9$ by $3$ gives $12x - 3y = 27$. The right side gets multiplied too.

Adding when you should subtract. Add when the coefficients are opposites ($3y$ and $-3y$). Subtract when they’re equal ($3x$ and $3x$). Adding $3x + 3x$ gives $6x$, which doesn’t eliminate anything.

Misreading 0 = 0 and 0 = 18. $0 = 0$ means infinitely many solutions, not “the answer is zero”. $0 = 18$ means no solution. Neither one means you made a mistake.

Forgetting the second variable. After you find $x$, back-substitute to find $y$, and give the answer as an ordered pair.

Practice

1. (Warm-up) Solve: $x + y = 10$ and $x - y = 4$.

Solution

Add the equations: $2x = 14$, so $x = 7$. Then $7 + y = 10$, so $y = 3$.

Check: $7 - 3 = 4$ ✓. The solution is $(7, 3)$.

2. (Warm-up) Solve: $3x + 2y = 13$ and $3x - 4y = 1$.

Solution

The $x$-coefficients are equal, so subtract the second equation from the first:

$$(3x + 2y) - (3x - 4y) = 13 - 1 \quad\Rightarrow\quad 6y = 12 \quad\Rightarrow\quad y = 2$$

Then $3x + 2(2) = 13$, so $3x = 9$ and $x = 3$.

Check: $3(3) - 4(2) = 1$ ✓. The solution is $(3, 2)$.

3. (Core) Solve: $5x + 2y = 4$ and $3x - y = 9$.

Solution

Multiply the second equation by $2$: $6x - 2y = 18$. Add it to the first:

$$11x = 22 \quad\Rightarrow\quad x = 2$$

Then $3(2) - y = 9$, so $y = -3$.

Check: $5(2) + 2(-3) = 10 - 6 = 4$ ✓. The solution is $(2, -3)$.

4. (Core) Solve: $4x + 3y = -1$ and $6x - 5y = 27$.

Solution

Eliminate $x$: the LCM of $4$ and $6$ is $12$. Multiply the first equation by $3$ and the second by $2$:

$$\begin{aligned} 12x + 9y &= -3 \\ 12x - 10y &= 54 \end{aligned}$$

Subtract the second from the first:

$$19y = -57 \quad\Rightarrow\quad y = -3$$

Then $4x + 3(-3) = -1$, so $4x = 8$ and $x = 2$.

Check: $6(2) - 5(-3) = 12 + 15 = 27$ ✓. The solution is $(2, -3)$.

5. (Core) Solve: $2y = 3x - 8$ and $5x + 4y = 6$.

Solution

Line up the first equation: $-3x + 2y = -8$. Multiply it by $2$ to get $-6x + 4y = -16$. Subtract this from the second equation:

$$(5x + 4y) - (-6x + 4y) = 6 - (-16) \quad\Rightarrow\quad 11x = 22 \quad\Rightarrow\quad x = 2$$

Then $2y = 3(2) - 8 = -2$, so $y = -1$.

Check: $5(2) + 4(-1) = 10 - 4 = 6$ ✓. The solution is $(2, -1)$.

6. (Core) Use elimination to decide how many solutions each system has.

• (a) $6x - 4y = 10$ and $3x - 2y = 5$
• (b) $6x - 4y = 10$ and $3x - 2y = 7$

Solution

(a) Multiply the second equation by $2$: $6x - 4y = 10$. Subtracting from the first gives $0 = 0$, which is always true. Infinitely many solutions: the equations describe the same line.

(b) Multiply the second equation by $2$: $6x - 4y = 14$. Subtracting from the first gives $0 = -4$, which is false. No solution: the lines are parallel.

7. (Core) Solve: $\dfrac{x}{3} + \dfrac{y}{2} = 4$ and $\dfrac{x}{2} - \dfrac{y}{4} = 2$.

Solution

Multiply the first equation by $6$ and the second by $4$:

$$\begin{aligned} 2x + 3y &= 24 \\ 2x - y &= 8 \end{aligned}$$

Subtract the second from the first: $4y = 16$, so $y = 4$. Then $2x - 4 = 8$, so $x = 6$.

Check in the original equations: $\dfrac{6}{3} + \dfrac{4}{2} = 2 + 2 = 4$ ✓ and $\dfrac{6}{2} - \dfrac{4}{4} = 3 - 1 = 2$ ✓. The solution is $(6, 4)$.

8. (Challenge) The point $(2, -1)$ is the solution of the system $ax + by = 5$ and $bx - ay = 5$. Find $a$ and $b$.

Solution

Substitute $x = 2$ and $y = -1$ into both equations:

$$\begin{aligned} 2a - b &= 5 \\ 2b + a &= 5 \end{aligned}$$

This is a new system, with $a$ and $b$ as the unknowns. Line it up as $2a - b = 5$ and $a + 2b = 5$. Multiply the first by $2$ to get $4a - 2b = 10$, then add the second:

$$5a = 15 \quad\Rightarrow\quad a = 3$$

Then $2(3) - b = 5$, so $b = 1$.

Check: the system is $3x + y = 5$ and $x - 3y = 5$. With $(2, -1)$: $6 - 1 = 5$ ✓ and $2 + 3 = 5$ ✓.

9. (Challenge) Simplify each equation first, then solve: $3(x + 2y) - 2(x - y) = 17$ and $4x - (x - 3y) = 9$.

Solution

Expand and collect like terms:

$$\begin{aligned} 3x + 6y - 2x + 2y &= 17 &&\Rightarrow\quad x + 8y = 17 \\ 4x - x + 3y &= 9 &&\Rightarrow\quad 3x + 3y = 9 \end{aligned}$$

Divide the second equation by $3$: $x + y = 3$. Subtract it from $x + 8y = 17$:

$$7y = 14 \quad\Rightarrow\quad y = 2$$

Then $x + 2 = 3$, so $x = 1$.

Check in the original equations: $3(1 + 4) - 2(1 - 2) = 15 + 2 = 17$ ✓ and $4(1) - (1 - 6) = 4 + 5 = 9$ ✓. The solution is $(1, 2)$.