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

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

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 25q + 5n = 125 100

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? 25q + 5n = 125 q + n = 13 25q + 5n = 125 -5q – 5n = -65 20q = 60 q = 3 There are 3 quarters and 10 nickels in the collection. 1 -5 Back substitution q + n = 13 3 + n = 13 n = 10

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 5n + 10d = 2500 100

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? 5n + 10d = 2500 n + d = 400 5n + 10d = 2500 -10n – 10d = -4000 -5n = -1500 n = 300 There are 300 nickels and 100 dimes in the collection. 1 -10 Back substitution n + d = 400 300 + d = 400 d = 100

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 100A + 75C = 37250 100

100A + 75C = 37250 A + C = 429 100A + 75C = 37250 -75A – 75C = -32175 25A = 5075 A = 203 They sold 203 adult tickets and 226 children tickets. 1 -75 Back substitution A + C = 429 203 + C = 429 C = 226 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?

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