Set up simultaneous equations from the total number of two types of items and their total price.
Example ①
Ten apples at 80 yen each and oranges at 120 yen each were bought, for a total of 10 items costing 1000 yen. How many of each were bought?
Solution: Let the number of apples be x and the number of oranges be y: { x+y=10、80x+120y=1000. Simplifying: { x+y=10、2x+3y=25. Elimination method: y=5,x=5。5 apples and 5 oranges
Ten apples at 80 yen each and oranges at 120 yen each were bought, for a total of 10 items costing 1000 yen. How many of each were bought?
Solution: Let the number of apples be x and the number of oranges be y: { x+y=10、80x+120y=1000. Simplifying: { x+y=10、2x+3y=25. Elimination method: y=5,x=5。5 apples and 5 oranges
Example ②
Admission is 300 yen per adult and 200 yen per child. There were 17 people, and the total was 3600 yen. Find the numbers of adults and children.
Solution: { x+y=17、300x+200y=3600 → { x+y=17、3x+2y=36. Elimination method: x=2,y=15。2 adults and 15 children
Admission is 300 yen per adult and 200 yen per child. There were 17 people, and the total was 3600 yen. Find the numbers of adults and children.
Solution: { x+y=17、300x+200y=3600 → { x+y=17、3x+2y=36. Elimination method: x=2,y=15。2 adults and 15 children
Key Points
- Set up two equations for the total number of items and the total price
- The price equation is the sum of unit price × quantity
- Large coefficients are easier to handle if both sides are divided by the same number
Practice Problem
- Twelve pencils at 50 yen each and ballpoint pens at 80 yen each were bought, for a total of 750 yen. How many of each were bought?
Answer and Explanation
- Answer:7 pencils and 5 ballpoint pens. ①x+y=12、②50x+80y=750（②÷10：5x+8y=75）。①×5：5x+5y=60。②−①×5：3y=15→y=5。From ①, x=7. Therefore, 7 pencils and 5 ballpoint pens.