SAT linear systems: tell one, none, or infinitely many solutions apart
Compare two equations before solving them. Their slopes and constants reveal whether the lines cross, stay parallel, or coincide.
SPSAT Practice Editorial5 min read
The short version
What you'll take away
- Different slopes mean one intersection and one solution.
- Equal slopes with different intercepts mean parallel lines and no solution.
- Equivalent equations describe the same line and have infinitely many solutions.
In this article
Classify the lines before you calculate
A solution to a system is a point that satisfies both equations. Two lines with different slopes meet once. Parallel lines never meet. If both equations name the same line, every point on it works.
Rewrite each equation as \(\displaystyle y=mx+b\) when that is quick. Compare the slope m first, then the intercept b. For vertical lines \(\displaystyle x=c\), compare the x-values c instead. A graph can confirm the result, but the equations give the exact answer.
See the skill
A two-check decision path
Compare slopes
Different slopes mean the lines cross once.
Different
One solution
Equal
Compare intercepts
Compare intercepts
Do this only when the slopes match.
Different
No solution: parallel lines
Equal
Infinitely many solutions: same line
Watch what elimination leaves behind
Consider \(\displaystyle 2x+y=7\) and \(\displaystyle 4x+2y=14\). Doubling the first equation produces the second, so the system has infinitely many solutions. In elimination, subtracting one equation from the other gives \(\displaystyle 0=0\), a true statement with no single x or y value.
Change the second equation to \(\displaystyle 4x+2y=15\). The left sides still have the same ratio, but the constants do not. Elimination now gives \(\displaystyle 0=1\), a contradiction, so there is no solution. A true identity signals the same line; a false statement signals parallel lines.
Try a one-step classification
Compare all coefficients, including the constant, before choosing.
Your turn
Quick practice
For what value of k does the system 3x − 2y = 5 and 6x − 4y = k have infinitely many solutions?
Hint: Multiply the entire first equation by 2.
Sources & further reading
One step from here
Put the idea to work.
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