Solving systems of equations with 2 variables

1. Solving systems of equations with 2 variables Word problems (Coins)

2. 8)A collection of quarters and nickels is worth $1.25. There are 13 coins in the collection. How many of each type of coin are there? Value of coins equation .25q + .05n = 1.25 Number of coins equation q + n = 13 100 25q + 5n = 125

3. 8)A collection of quarters and nickels is worth $1.25. There are 13 coins in the collection. How many of each type of coin are there? 1 25q + 5n = 125 q + n = 13 25q + 5n = 125 -5q – 5n = -65 20q = 60 q = 3 -5 Back substitution q + n = 13 3 + n = 13 n = 10 There are 3 quarters and 10 nickels in the collection.

4. 9)A collection of nickels and dimes is worth $25. There are 400 coins in the collection. How many of each type of coin are there? Value of coins equation .05n + .10d = 25 Number of coins equation q + n = 400 100 5n + 10d = 2500

5. 9)A collection of nickels and dimes is worth $25. There are 400 coins in the collection. How many of each type of coin are there? 1 5n + 10d = 2500 n + d = 400 5n + 10d = 2500 -10n – 10d = -4000 -5n = -1500 n = 300 -10 Back substitution n + d = 400 300 + d = 400 d = 100 There are 300 nickels and 100 dimes in the collection.

6. 10)There are 429 people at a play. Admission is $1 for adults and 75 cents for children. The receipts were $372.50. How many adults and children tickets were sold? Value of tickets equation 1A + .75C = 372.50 Number of tickets equation A + C = 429 100 100A + 75C = 37250

7. 10)There are 429 people at a play. Admission is $1 for adults and 75 cents for children. The receipts were $372.50. How many adults and children tickets were sold? 1 100A + 75C = 37250 A + C = 429 100A + 75C = 37250 -75A – 75C = -32175 25A = 5075 A = 203 -75 Back substitution A + C = 429 203 + C = 429 C = 226 They sold 203 adult tickets and 226 children tickets.