Solving Systems Algebraically
Graphing systems is visual and helps you understand what's happening geometrically, but it's not always precise or practical. Algebraic methods—substitution and elimination—let you find exact solutions without drawing a graph.
Both methods work by reducing a two-variable system to a single equation with one variable, which you can then solve using techniques you already know.
The Substitution Method
Substitution works well when one equation is already solved for a variable (or can easily be solved for a variable).
The strategy: solve one equation for one variable, then substitute that expression into the other equation.
Example: Solve using substitution.
$$\begin{cases} y = 2x + 1 \\ 3x + y = 11\end{cases}$$
The first equation is already solved for \(y\). Substitute \(2x + 1\) for \(y\) in the second equation:
$$3x + (2x + 1) = 11$$
Now we have one equation with one variable:
$$5x + 1 = 11$$ $$5x = 10$$ $$x = 2$$
Now that we know \(x = 2\), substitute it back into either original equation to find \(y\). Let's use the first one:
$$y = 2(2) + 1 = 5$$
Solution: \((2, 5)\)
Check in both equations: \(5 = 2(2) + 1 = 5\) ✓ and \(3(2) + 5 = 6 + 5 = 11\) ✓
Example: Solve using substitution.
$$\begin{cases} x + y = 8 \\ 2x - y = 1\end{cases}$$
Neither equation is solved for a variable, but the first one is easy to solve for \(y\):
$$y = 8 - x$$
Substitute into the second equation:
$$2x - (8 - x) = 1$$ $$2x - 8 + x = 1$$ $$3x - 8 = 1$$ $$3x = 9$$ $$x = 3$$
Find \(y\): $$y = 8 - 3 = 5$$
Solution: \((3, 5)\)
The Elimination Method
Elimination (also called addition method) works by adding or subtracting the equations to eliminate one variable.
The strategy: manipulate the equations so that when you add them together, one variable cancels out.
Example: Solve using elimination.
$$\begin{cases} x + y = 7 \\ x - y = 3\end{cases}$$
Notice that the \(y\) terms are already opposites (\(+y\) and \(-y\)). If we add the equations, \(y\) will cancel:
$$\begin{align} x + y &= 7 \
- (x - y &= 3) \ \hline 2x &= 10 \end{align}$$
$$x = 5$$
Substitute \(x = 5\) into either original equation: $$5 + y = 7$$ $$y = 2$$
Solution: \((5, 2)\)
Example: Solve using elimination.
$$\begin{cases} 3x + 2y = 12 \\ 5x - 2y = 4\end{cases}$$
The \(y\) terms are opposites, so add:
$$\begin{align} 3x + 2y &= 12 \
- (5x - 2y &= 4) \ \hline 8x &= 16 \end{align}$$
$$x = 2$$
Find \(y\): $$3(2) + 2y = 12$$ $$6 + 2y = 12$$ $$2y = 6$$ $$y = 3$$
Solution: \((2, 3)\)
When Coefficients Aren't Opposites
Often, you need to multiply one or both equations by a constant to make coefficients opposites.
Example: Solve using elimination.
$$\begin{cases} 2x + 3y = 13 \\ x + y = 5\end{cases}$$
We can eliminate \(x\) by multiplying the second equation by -2:
$$\begin{cases} 2x + 3y = 13 \\ -2x - 2y = -10\end{cases}$$
Now add:
$$\begin{align} 2x + 3y &= 13 \
- (-2x - 2y &= -10) \ \hline y &= 3 \end{align}$$
Find \(x\): $$x + 3 = 5$$ $$x = 2$$
Solution: \((2, 3)\)
Example: Solve using elimination.
$$\begin{cases} 3x + 4y = 10 \\ 2x + 3y = 7\end{cases}$$
Let's eliminate \(x\). We need the coefficients of \(x\) to be opposites. The LCM of 3 and 2 is 6, so multiply the first equation by 2 and the second by -3:
$$\begin{cases} 6x + 8y = 20 \\ -6x - 9y = -21\end{cases}$$
Add:
$$\begin{align} 6x + 8y &= 20 \
- (-6x - 9y &= -21) \ \hline -y &= -1 \end{align}$$
$$y = 1$$
Find \(x\): $$3x + 4(1) = 10$$ $$3x + 4 = 10$$ $$3x = 6$$ $$x = 2$$
Solution: \((2, 1)\)
Choosing Between Substitution and Elimination
Use substitution when:
- One equation is already solved for a variable
- One equation can easily be solved for a variable
- You see something like \(y = \text{something}\) or \(x = \text{something}\)
Use elimination when:
- Coefficients are already opposites or can easily be made opposites
- Both equations are in standard form
- Neither equation is easily solved for a single variable
Some systems work equally well with either method—choose whichever feels more natural to you.
Special Cases Revisited
No solution: When you solve and get a false statement like \(0 = 5\).
Example: $$\begin{cases} 2x + y = 4 \\ 2x + y = 7\end{cases}$$
Subtract the first from the second: $$0 = 3$$
This is false, so no solution exists. The lines are parallel.
Infinitely many solutions: When you get a true statement like \(0 = 0\).
Example: $$\begin{cases} x + 2y = 6 \\ 2x + 4y = 12\end{cases}$$
Multiply the first by -2: $$\begin{cases} -2x - 4y = -12 \\ 2x + 4y = 12\end{cases}$$
Add: $$0 = 0$$
This is always true, meaning the equations represent the same line. Infinitely many solutions.
Give These a Try
Solve each system using either substitution or elimination:
$$\begin{cases} y = 3x - 5 \\ 2x + y = 10\end{cases}$$
$$\begin{cases} x + y = 12 \\ x - y = 4\end{cases}$$
$$\begin{cases} 3x + 2y = 16 \\ 2x - y = 4\end{cases}$$
$$\begin{cases} 4x + 3y = 18 \\ 2x + y = 8\end{cases}$$
Solutions:
-
Substitution: \(2x + (3x - 5) = 10\), so \(5x = 15\), \(x = 3\), \(y = 4\). Solution: \((3, 4)\)
-
Elimination: add equations to get \(2x = 16\), \(x = 8\), \(y = 4\). Solution: \((8, 4)\)
-
Elimination or substitution. From second equation: \(y = 2x - 4\). Substitute: \(3x + 2(2x - 4) = 16\), so \(7x = 24\), \(x = \frac{24}{7}\), \(y = \frac{20}{7}\). Solution: \((\frac{24}{7}, \frac{20}{7})\)
-
Elimination: multiply second by -3: \(-6x - 3y = -24\). Add to first: \(-2x = -6\), \(x = 3\), \(y = 2\). Solution: \((3, 2)\)
Things That Trip Students Up
In substitution, forgetting parentheses when you plug in an expression is a recipe for disaster. If \(y = 2x + 1\) and you're substituting into \(3x - y = 5\), write \(3x - (2x + 1) = 5\), not \(3x - 2x + 1 = 5\). Those parentheses matter because of the minus sign.
In elimination, adding or subtracting incorrectly usually comes from not lining things up properly or rushing through the arithmetic. Line up your equations carefully, write out each step, and watch those signs.
Multiplying only one side of an equation when setting up elimination defeats the entire purpose. When you multiply an equation by a constant, multiply every single term on both sides. If you're multiplying \(2x + y = 5\) by 3, you get \(6x + 3y = 15\), not \(6x + 3y = 5\).
Not checking the solution in both original equations is lazy and costly. Your answer might satisfy one equation perfectly but fail the other, which means it's wrong. Check both.
Stopping after finding just \(x\) or just \(y\) leaves the problem incomplete. You need both variables, written as an ordered pair \((x, y)\). Don't stop halfway.