What this lesson covers
- The two ledgers every two-purchase story is really made of
- Why the count total and the money total never share a row
- Folding one equation into the other, and checking both rows at the end
- The swap trick: pretend they were all the expensive one, then count the overshoot
- Why the number you solved for is often not the number they asked for
Chapters
- 0:00Ten tickets, ninety four dollars
- 0:04Two prices, one evening
- 0:17The setup: which numbers are counts
- 0:39Two rows, two equations, one fold
- 1:33The no-algebra swap trick
- 2:09Your turn
- 2:51Three ways this goes wrong
- 3:13Counts and cash
Lesson transcript
The narration of the video, word for word, under its chapter headings.
0:17The setup: which numbers are counts
Adult tickets are eleven dollars. Child tickets are seven. Ten tickets sold, ninety four dollars collected. How many were child tickets? Every number in that story belongs to one of exactly two ledgers, and the wrong answers all come from filing one of them wrong.
0:39Two rows, two equations, one fold
Ledger one counts tickets. a plus c equals ten. Ledger two counts dollars. Eleven a plus seven c equals ninety four. The ten is a count. The ninety four is money. Neither ever visits the other row. Now fold one into the other. The count row says a equals ten minus c. Substitute. Eleven times ten minus c, plus seven c, equals ninety four. One hundred ten minus four c equals ninety four. Four c equals sixteen. c equals four. Four child tickets, six adult. And check both ledgers. Six plus four is ten. Sixty six plus twenty eight is ninety four. Both rows close. The answer is A.
1:33The no-algebra swap trick
Here is the no algebra version. Pretend all ten tickets were adult. That is one hundred ten dollars, sixteen too much. Each child swap saves four dollars. Sixteen over four, four swaps. Four child tickets. Same answer, zero substitution. And watch choice C. Six is the number of adult tickets. Solve the system perfectly, answer the wrong variable, lose the point. The question said child. Circle the variable they asked for before you solve.
2:09Your turn
Pens cost two dollars. Notebooks cost three. Maya buys fourteen items for thirty four dollars. How many notebooks? Counts. p plus n is fourteen. Cash. Two p plus three n is thirty four. Substitute. Twenty eight plus n equals thirty four. n equals six. Six notebooks, eight pens. Sixteen plus eighteen is thirty four. Both ledgers close. The answer is A.
2:51Three ways this goes wrong
Three ways this goes wrong. Answering the other variable, the system was right and the point is gone. Swapping the prices, seven a plus eleven c is a different story. And mixing ledgers, subtracting fourteen items from thirty four dollars gives ten of nothing. Counts and cash never share a row.
3:13Counts and cash
Two ledgers. One for counts, one for cash. Fold, solve, check both.
Read the written version: the Systems of two linear equations strategy guide, then try 25 hard Systems of two linear equations questions with full explanations. Both are free, no account needed.