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.
Click to begin the narrated lesson
Solve systems of two linear equations in two variables algebraically