Two conditions at once
What you'll learn
See a system's solution as the point where two lines cross — and why some systems have no solution or infinitely many.
One equation with two unknowns can't be pinned down: x + y = 10 is satisfied by (1, 9), (4, 6), (−3, 13)… a whole line of answers. To lock onto a single point you need a second condition. Two equations, considered together, are a system — and solving it means finding the points that satisfy both.
The picture
Each equation draws a line: all the points that satisfy it. A point satisfying both equations must sit on both lines — so the solutions of the system are exactly the places the lines meet. Three things can happen:
| Lines | Solutions | How to spot it from the equations |
|---|---|---|
| cross | exactly one | different slopes |
| parallel | none | same slope, different intercepts |
| identical | infinitely many | one equation is a multiple of the other |
Reading the fate before solving
Put both equations in y = mx + b form and compare:
- y = 2x + 1 and y = −x + 4 → slopes 2 and −1 differ → one solution.
- y = 3x − 2 and y = 3x + 5 → same slope, different b → no solution. The system demands a number that's simultaneously 3x − 2 and 3x + 5 — impossible.
- 2x + y = 4 and 4x + 2y = 8 → the second is the first times 2 → same line, infinitely many solutions.
Checking a claimed solution
Is (3, 1) a solution of x + y = 4 and 2x − y = 5? Feed it to both: 3 + 1 = 4 ✓ and 6 − 1 = 5 ✓. Both hold, so yes. One equation passing and one failing means no — a system's solution must satisfy every equation, not most of them.
Why this matters
"Where do two conditions hold at once?" is one of math's most repeated questions — break-even points in business, matching supply with demand, finding where two moving objects meet. Next lesson: how to find the crossing point exactly, without graph paper.
Check your understanding
Question 1 of 2
Graphically, the solution of a 2-variable system is: