In this lesson:
By the end of this lesson, you should be able to:
Decide by sight, before any algebra:
One has no solution, one has infinitely many. Which is which?
The expression cannot equal both 5 and 6.
So no pair can satisfy both.
Same slope, different intercepts — they never cross.
Divide the second by 2: it becomes — the same line.
Solve and :
Exactly one solution: .
The ticket system from Lesson 2 was , .
Graphed, the lines cross at — confirming 3 adults, 5 children.
How many solutions? Decide by inspection:
Is one equation a multiple of the other?
One system can be approached three ways. Match each to its best route:
Inspect, then pick the cheapest method, then verify.
Solve and by substitution:
Classify each by inspection — do not solve:
For each system, name the method you'd use and why:
Match the form to the method.
For and , decide on your own:
One, none, or infinitely many? Write your verdict and one reason.
For each, classify the number of solutions. For the one with a single solution, solve it and verify:
means no solution; means infinitely many A fraction or negative is a valid solution Inspect first — don't grind a special case
In Lesson 3, the systems arrive hidden inside word problems. You'll build the equations yourself, then solve, interpret, and check that your answer makes sense.