In this lesson:
By the end of this lesson, you should be able to:
A group buys 8 tickets for $57. Adult tickets cost $9, child tickets $6.
Two unknowns at once — guessing is slow.
From , we can write:
We just expressed one unknown using the other.
Substitute into :
3 adult tickets and 5 child tickets.
Solve and :
Each step has a reason — not a recipe to memorize.
Solve and . We've started:
In and , which variable is cheapest to isolate, and from which equation?
Think about coefficients of 1.
Solve and .
Isolating means fractions. But notice: both have .
What if we just subtract one equation from the other?
For and , you could instead match the terms.
Different multipliers, same answer .
Hint: multiply the first equation by 3 to match the terms, then subtract.
A student subtracts from and writes .
The cancels correctly — but what about the terms?
For and , decide on your own:
Which method? Which variable? Why? Write it down before solving.
Solve and from start to finish.
Find both values, then verify in both equations.
Distribute the minus to every term when subtracting Found ? You're not done — find too Substitute into the other equation, not the same one
In Lesson 3, you'll add a third skill — solving simple systems by inspection — and a guide for choosing the fastest method for any system.