Special Cases: No Solution and Infinitely Many Solutions
Algebraically and graphically identify systems with no solution (parallel lines, contradiction) and infinitely many solutions (coincident lines, identity). Interpret these in real-world contexts.
A lesson of Algebra 1, in Systems of Equations & Inequalities, about 25 minutes.
When There Is No Solution, or Too Many
By the end of this lesson, you will be able to recognize a system with no solution or infinitely many, tell them apart from the algebra as well as from the graph, and write the answer properly.
Both Variables Disappear
Solve a normal system and you get a value. Solve one of these and something else happens: the variables cancel and leave a statement with no letters in it.
$$\begin{cases} 2x + y = 5 \ 4x + 2y = 3 \end{cases}$$
Multiply the first by $2$ and subtract:
$$4x + 2y = 10$$ $$4x + 2y = 3$$ $$0 = 7$$
No $x$, no $y$, and a statement that is simply false. Nothing you choose for $x$ and $y$ can make $0$ equal $7$, so there is no solution.
$$\begin{cases} 3x - y = 4 \ 6x - 2y = 8 \end{cases}$$
Multiply the first by $2$ and subtract:
$$0 = 0$$
True, and true whatever $x$ and $y$ are. Every point on the line satisfies both equations, so there are infinitely many solutions.
The Test
When the variables vanish, read what is left.
| What remains | Meaning | Answer |
|---|---|---|
| A false statement, like $0 = 7$ | The conditions contradict | No solution |
| A true statement, like $0 = 0$ | The conditions are the same | Infinitely many |
That is the whole rule, and you have met it before. In the very first unit, solving $3x + 5 = 3x - 2$ gave $5 = -2$ and the answer was "no solution". This is the same event with two variables instead of one.
What it does NOT mean. Getting $x = 0$ is an ordinary answer, not a special case. The special cases are when the VARIABLE disappears, not when it equals zero, and confusing those two is the most common error here.
Example 1: Reading the Leftover
Here is the whole method on one system, worked all the way through. $$\begin{cases} x + 3y = 6 \ 2x + 6y = 5 \end{cases}$$
- Multiply the first equation by 2. Write it 2x + 6y = 12
- Subtract the second from that. What is left on the left side? 0
- What is left on the right side? 7