Solving Systems Algebraically
Graphing helps you understand what's happening geometrically, but it isn't always precise or practical. The two algebraic methods covered here, substitution and elimination, give 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:
$$3x + (2x + 1) = 11$$
That leaves one equation with one variable:
$$5x + 1 = 11$$ $$5x = 10$$ $$x = 2$$
With \(x = 2\) in hand, substitute it back into either original equation to find \(y\). Using the first equation:
$$y = 2(2) + 1 = 5$$
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: \((3, 5)\)
Quick check: verify that \((3, 5)\) satisfies both original equations. Show answerFirst equation: \(3 + 5 = 8\) ✓. Second equation: \(2(3) - 5 = 6 - 5 = 1\) ✓. Both check out.
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 \(-y\)). Adding the equations will cancel \(y\):
$$\begin{align} x + y &= 7 \
- (x - y &= 3) \ \hline 2x &= 10 \end{align}$$
$$x = 5$$
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.
$$\begin{cases} 2x + 3y = 13 \\ x + y = 5\end{cases}$$
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.
$$\begin{cases} 3x + 4y = 10 \\ 2x + 3y = 7\end{cases}$$
Eliminate \(x\). The coefficients of \(x\) need to be opposites, and 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. Pick whichever feels more natural for the problem in front of 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:
$$\begin{cases} y = 3x - 5 \\ 2x + y = 10\end{cases}$$ Show answerSubstitution: replace \(y\) with \(3x - 5\) in the second equation: \(2x + (3x - 5) = 10\), so \(5x = 15\), \(x = 3\), then \(y = 4\). Solution: \((3, 4)\)
$$\begin{cases} x + y = 12 \\ x - y = 4\end{cases}$$ Show answerElimination: add the equations to get \(2x = 16\), so \(x = 8\), then \(y = 4\). Solution: \((8, 4)\)
$$\begin{cases} 3x + 2y = 16 \\ 2x - y = 4\end{cases}$$ Show answerSubstitution works well here. From the second equation: \(y = 2x - 4\). Substitute: \(3x + 2(2x - 4) = 16\), so \(7x = 24\), \(x = \frac{24}{7}\), \(y = \frac{20}{7}\). Solution: \(\left(\frac{24}{7}, \frac{20}{7}\right)\)
$$\begin{cases} 4x + 3y = 18 \\ 2x + y = 8\end{cases}$$ Show answerElimination: multiply the second equation by -3: \(-6x - 3y = -24\). Add to the first equation: \(-2x = -6\), so \(x = 3\), then \(y = 2\). Solution: \((3, 2)\)
What's Next?
A few habits that keep these problems clean. In substitution, parentheses around the substituted expression matter — when \(y = 2x + 1\) gets plugged into \(3x - y = 5\), it becomes \(3x - (2x + 1) = 5\), not \(3x - 2x + 1 = 5\); the minus sign in front has to distribute. In elimination, when you multiply an equation by a constant, multiply every term on both sides; \(2x + y = 5\) times 3 is \(6x + 3y = 15\), not \(6x + 3y = 5\). Always check both original equations with your solution — a wrong answer often satisfies one but not the other. And don't stop after finding just \(x\) or just \(y\); a system solution is an ordered pair \((x, y)\).
The next step is systems with three or more variables, which uses the same elimination idea but with more bookkeeping.