MISSION · Algebra
Systems of Two Equations
Where two lines meet, and what it means when they never do
LEVEL 1 · EXPLORE
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LEVEL 2 · LEARN
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The idea in one sentence
Two equations in two unknowns are a single question: which pair of values makes both true at the same time?
The picture first
Each equation, on its own, has infinitely many solutions — every point on its line. The system asks for the points that lie on both lines, which is normally exactly one: the intersection.
That geometric picture predicts the three possible outcomes before you calculate anything:
- The lines cross once — one solution.
- The lines are parallel — no solution. The equations contradict each other.
- The lines are identical — infinitely many solutions. The second equation was the first one in disguise.
If your algebra ends at something absurd like \(0 = 5\), you’ve found parallel lines. If it ends at something vacuous like \(0 = 0\), the two equations were the same line.
Substitution
Rearrange one equation for one variable, then put that expression into the other.
\[y = 2x + 1 \qquad 3x + y = 11\]The first already gives \(y\), so substitute it: \(3x + (2x + 1) = 11\), so \(5x = 10\) and \(x = 2\). Then back into the first: \(y = 5\). The solution is \((2, 5)\).
Best when one variable is already alone, or easy to isolate.
Elimination
Add or subtract the equations so one variable cancels.
\[3x + 2y = 18 \qquad 3x - y = 9\]Both start with \(3x\), so subtracting the second from the first removes it entirely: \(3y = 9\), giving \(y = 3\). Put that back into either equation and \(3x = 12\), so \(x = 4\).
When the coefficients don’t already match, scale one equation until they do — multiplying every term in it, on both sides.
Best when the coefficients already line up, or almost do.
Both methods give the same answer. Pick by looking at the equations, not by habit.
Always check both
A solution has to satisfy both equations. Substituting back into only the one you used last will confirm an arithmetic slip rather than catch it. Two checks cost fifteen seconds.
Common mix-ups
- Solving for one variable and stopping. The answer is a pair.
- Scaling only part of an equation. Multiply every term, including the constant.
- Adding when you should subtract, which doubles a variable instead of cancelling it.
- Reading no solution as zero. “No solution” means no pair works at all; \(x = 0\) is a perfectly good solution.
Check yourself
Solve \(y = x + 4\) and \(2x + y = 10\), then say what would have happened if the second equation had been \(y = x - 3\).
Substituting gives \(2x + x + 4 = 10\), so \(x = 2\) and \(y = 6\). With \(y = x - 3\) the two lines would both have slope 1 and never meet — parallel, no solution.
LEVEL 3 · REFLECT
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