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Published byReed Nelson Modified about 1 year ago

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

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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 =

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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 = q – 5n = q = 60 q = 3 There are 3 quarters and 10 nickels in the collection Back substitution q + n = n = 13 n = 10

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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 =

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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 = n – 10d = n = n = 300 There are 300 nickels and 100 dimes in the collection Back substitution n + d = d = 400 d = 100

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

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

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