Table of Contents
The elimination method is a systematic way to solve a system of linear equations by combining the equations to remove (eliminate) one of the variables. Once one variable is eliminated, the system becomes easier to solve.
In this chapter we will:
- Focus on the mechanics of elimination.
- Practice choosing which variable to eliminate.
- Handle common special cases and pitfalls.
The general ideas of systems of linear equations and their solutions are assumed from the parent chapters.
Basic idea of elimination
Consider a system of two linear equations with two variables:
$$
\begin{cases}
a_1x + b_1y = c_1 \\
a_2x + b_2y = c_2
\end{cases}
$$
The idea of elimination:
- Multiply one or both equations (if needed) so that the coefficient of one variable is the same (or opposite) in both equations.
- Add or subtract the equations to eliminate that variable.
- Solve the resulting single-variable equation.
- Substitute back to find the other variable.
- Check the solution in the original equations.
The key operation is: combine the equations so that one variable disappears.
Elimination when coefficients already match
Sometimes, one variable already has coefficients that are easy to eliminate.
Example:
$$
\begin{cases}
2x + 3y = 13 \\
2x - y = 5
\end{cases}
$$
Here, the $x$-terms both have coefficient $2$. If we subtract the second equation from the first, the $x$-terms will cancel.
Step 1: Decide how to combine the equations
We want to eliminate $x$:
- $2x - 2x = 0$
So compute:
$$
(2x + 3y) - (2x - y) = 13 - 5
$$
Step 2: Carry out the subtraction
On the left:
- $2x - 2x = 0$
- $3y - (-y) = 3y + y = 4y$
So:
$$
4y = 8
$$
Step 3: Solve for $y$
$$
y = \frac{8}{4} = 2
$$
Step 4: Substitute back into either original equation
Use $2x - y = 5$:
$$
2x - 2 = 5 \\
2x = 7 \\
x = \frac{7}{2}
$$
Solution: $x = \frac{7}{2}$, $y = 2$.
The solution can be written as the ordered pair $\left(\frac{7}{2}, 2\right)$.
Elimination when one coefficient is the negative of the other
If one coefficient is the negative of the other, adding the equations will eliminate that variable.
Example:
$$
\begin{cases}
3x + 4y = 1 \\
-3x + 2y = 11
\end{cases}
$$
Here the $x$-coefficients are $3$ and $-3$.
Step 1: Add the equations
Add left sides and right sides:
$$
(3x + 4y) + (-3x + 2y) = 1 + 11
$$
On the left:
- $3x + (-3x) = 0$
- $4y + 2y = 6y$
So:
$$
6y = 12
$$
Step 2: Solve for $y$
$$
y = \frac{12}{6} = 2
$$
Step 3: Substitute back
Use $3x + 4y = 1$:
$$
3x + 4(2) = 1 \\
3x + 8 = 1 \\
3x = -7 \\
x = -\frac{7}{3}
$$
Solution: $x = -\frac{7}{3}$, $y = 2$.
Elimination when coefficients do not match
Often, the coefficients are not ready for elimination. Then you multiply one or both equations by a number to create matching (or opposite) coefficients.
Multiplying one equation only
Example:
$$
\begin{cases}
x + 2y = 9 \\
3x + 2y = 15
\end{cases}
$$
Here, the $y$-coefficients already match ($2$ and $2$). We can eliminate $y$ directly by subtracting, but we can also see how multiplying can help in other cases. Let’s eliminate $x$ instead.
The $x$-coefficients are $1$ and $3$. To eliminate $x$, we can make them opposites, for instance $3$ and $-3$.
Multiply the first equation by $-3$:
$$
-3(x + 2y) = -3(9)
$$
which gives:
$$
-3x - 6y = -27
$$
Now our system is:
$$
\begin{cases}
-3x - 6y = -27 \\
3x + 2y = 15
\end{cases}
$$
Add the equations:
$$
(-3x - 6y) + (3x + 2y) = -27 + 15
$$
On the left:
- $-3x + 3x = 0$
- $-6y + 2y = -4y$
So:
$$
-4y = -12
$$
and
$$
y = 3
$$
Substitute into $x + 2y = 9$:
$$
x + 2(3) = 9 \\
x + 6 = 9 \\
x = 3
$$
Solution: $(3, 3)$.
Multiplying both equations
Sometimes you need to change both equations to create matching coefficients.
Example:
$$
\begin{cases}
2x + 3y = 7 \\
5x - 4y = 3
\end{cases}
$$
Neither $2$ and $5$ nor $3$ and $-4$ are already multiples of each other. Pick a variable to eliminate. Let’s eliminate $x$.
The $x$-coefficients are $2$ and $5$. Their least common multiple (LCM) is $10$. We can create $10x$ and $-10x$.
- Multiply the first equation by $5$:
$$
5(2x + 3y) = 5(7) \
10x + 15y = 35
$$ - Multiply the second equation by $-2$:
$$
-2(5x - 4y) = -2(3) \
-10x + 8y = -6
$$
Now we have:
$$
\begin{cases}
10x + 15y = 35 \\
-10x + 8y = -6
\end{cases}
$$
Add the equations:
$$
(10x + 15y) + (-10x + 8y) = 35 + (-6)
$$
On the left:
- $10x - 10x = 0$
- $15y + 8y = 23y$
So:
$$
23y = 29
$$
and
$$
y = \frac{29}{23}
$$
Substitute into one original equation, for instance $2x + 3y = 7$:
$$
2x + 3\left(\frac{29}{23}\right) = 7
$$
Compute:
$$
2x + \frac{87}{23} = 7
$$
Write $7$ with denominator $23$:
$$
7 = \frac{161}{23}
$$
So:
$$
2x + \frac{87}{23} = \frac{161}{23} \\
2x = \frac{161}{23} - \frac{87}{23} = \frac{74}{23}
$$
Thus:
$$
x = \frac{74}{46} = \frac{37}{23}
$$
Solution: $\left(\frac{37}{23}, \frac{29}{23}\right)$.
Choosing the best variable to eliminate
You are free to choose which variable to eliminate. A good choice can make the work much easier.
Tips:
- If one variable already has matching coefficients (like $2y$ and $2y$), eliminate that one.
- If one variable has a coefficient of $1$ or $-1$, that variable is often a good candidate, especially if you might switch to substitution.
- Look for the smallest common multiples to avoid large numbers.
- You can also choose based on which will avoid fractions, if possible.
Example:
$$
\begin{cases}
4x + 5y = 1 \\
6x - 15y = 9
\end{cases}
$$
Compare eliminating $x$ vs $y$:
- For $x$: coefficients $4$ and $6$ → LCM $12$.
- For $y$: coefficients $5$ and $-15$ → LCM $15$.
But notice: $5$ and $-15$ are already multiples; simply multiply the first equation by $3$:
$$
3(4x + 5y) = 3(1) \\
12x + 15y = 3
$$
Now the system is:
$$
\begin{cases}
12x + 15y = 3 \\
6x - 15y = 9
\end{cases}
$$
Add:
$$
(12x + 15y) + (6x - 15y) = 3 + 9 \\
18x = 12 \\
x = \frac{12}{18} = \frac{2}{3}
$$
Then substitute to find $y$. In this case, eliminating $y$ was more straightforward.
Using subtraction vs addition
You can use either addition or subtraction to eliminate a variable, depending on the signs.
- If coefficients are equal (like $4y$ and $4y$), subtract the equations to get $0$.
- If coefficients are opposites (like $4y$ and $-4y$), add the equations to get $0$.
You can also convert one situation into the other by multiplying an equation by $-1$.
Example:
$$
\begin{cases}
x + 4y = 10 \\
3x + 4y = 2
\end{cases}
$$
To eliminate $y$:
- Coefficients are $4$ and $4$ (same sign).
- Subtract the first equation from the second:
$$
(3x + 4y) - (x + 4y) = 2 - 10
$$
On the left:
- $3x - x = 2x$
- $4y - 4y = 0$
So:
$$
2x = -8 \Rightarrow x = -4
$$
Then substitute back to find $y$.
Alternatively, multiply the first equation by $-1$:
$$
-x - 4y = -10
$$
and add to the second equation. The result is the same.
Fractions in elimination
Fractions often appear in systems of equations. You can still do elimination, but it is usually easier if you clear denominators first.
Example:
$$
\begin{cases}
\frac{1}{2}x + \frac{1}{3}y = 4 \\
\frac{1}{4}x - \frac{1}{6}y = 1
\end{cases}
$$
Step 1: Clear denominators in both equations.
For the first equation, denominators are $2$ and $3$. LCM is $6$:
$$
6\left(\frac{1}{2}x + \frac{1}{3}y\right) = 6 \cdot 4 \\
3x + 2y = 24
$$
For the second equation, denominators are $4$ and $6$. LCM is $12$:
$$
12\left(\frac{1}{4}x - \frac{1}{6}y\right) = 12 \cdot 1 \\
3x - 2y = 12
$$
Now the system is:
$$
\begin{cases}
3x + 2y = 24 \\
3x - 2y = 12
\end{cases}
$$
Step 2: Eliminate $y$ by adding:
$$
(3x + 2y) + (3x - 2y) = 24 + 12 \\
6x = 36 \Rightarrow x = 6
$$
Step 3: Substitute back to find $y$:
$$
3(6) + 2y = 24 \\
18 + 2y = 24 \\
2y = 6 \Rightarrow y = 3
$$
Solution: $(6, 3)$.
The main strategy: multiply to remove fractions, then use elimination as usual.
Special cases: no solution and infinitely many solutions
When you use elimination, sometimes all variables disappear, leaving either a true or a false statement. This tells you about the type of solution.
No solution (inconsistent system)
If elimination leads to a false statement like $0 = 5$, the system has no solution; the lines are parallel and distinct.
Example:
$$
\begin{cases}
2x + 4y = 6 \\
x + 2y = 10
\end{cases}
$$
Eliminate $x$. Multiply the second equation by $2$:
$$
2(x + 2y) = 2(10) \\
2x + 4y = 20
$$
Now we have:
$$
\begin{cases}
2x + 4y = 6 \\
2x + 4y = 20
\end{cases}
$$
Subtract the first from the second:
$$
(2x + 4y) - (2x + 4y) = 20 - 6 \\
0 = 14
$$
This is impossible. So there is no solution.
Infinitely many solutions (dependent system)
If elimination leads to a true statement like $0 = 0$, and both equations are actually the same line, there are infinitely many solutions.
Example:
$$
\begin{cases}
3x - 9y = 12 \\
x - 3y = 4
\end{cases}
$$
Multiply the second equation by $3$:
$$
3(x - 3y) = 3 \cdot 4 \\
3x - 9y = 12
$$
Now both equations are identical:
$$
\begin{cases}
3x - 9y = 12 \\
3x - 9y = 12
\end{cases}
$$
Subtract one from the other:
$$
0 = 0
$$
This is always true, and it does not allow you to solve for a particular $x$ and $y$. Instead, there are infinitely many solutions: every point on the line $x - 3y = 4$ is a solution.
Summary of these cases:
- If all variables eliminate and you get a false statement (like $0 = 5$), the system has no solution.
- If all variables eliminate and you get a true statement (like $0 = 0$), and the equations are multiples of each other, there are infinitely many solutions.
Elimination in word problems (brief outline)
Detailed translation of word problems into equations is handled in the word-problems chapter. Here is how elimination fits in once you have the system.
Example (structure only):
Suppose a word problem leads to:
$$
\begin{cases}
x + y = 23 \\
2x - y = 7
\end{cases}
$$
Use elimination:
Add the equations:
$$
(x + y) + (2x - y) = 23 + 7 \\
3x = 30 \Rightarrow x = 10
$$
Then substitute into $x + y = 23$:
$$
10 + y = 23 \Rightarrow y = 13
$$
After solving, you interpret $x$ and $y$ back in the context of the problem (for example, numbers of items, prices, ages, etc.).
The important part here is: once you have a system of equations from the situation, elimination is often a clean and structured way to find the values.
Common mistakes and how to avoid them
- Not multiplying the entire equation:
When you multiply an equation, every term on both sides must be multiplied, including the constant term on the right. - Sign errors when subtracting:
Subtracting an equation means subtracting every term, not just some. Writing the second equation in parentheses and distributing the negative sign can help:
$$
(2x + 3y) - (x - 4y) = (2x + 3y) + (-x + 4y)
$$ - Forgetting to check the solution:
Substitution back into both original equations catches arithmetic mistakes early. - Choosing a complicated elimination path:
Before starting, look for the simplest variable to eliminate: matching or opposite coefficients, or small numbers.
By practicing a variety of systems and deliberately choosing how to eliminate, you will become faster and more accurate with this method.