Pre-Algebra Unit 6 Review Unit 6 Review 1) Solve by graphing. 3π₯ + 4π¦ = 16 1 π¦ = π₯ β1 2 Unit 6 Review 1) Solve by graphing. 3π₯ + 4π¦ = 16 1 π¦ = π₯ β1 2 (4,1) Unit 6 Review 2. Solve by substitution. 4π₯ + 6π¦ = 10 2π¦ + π₯ = 2 Unit 6 Review 2. Solve by substitution. 4π₯ + 6π¦ = 10 2π¦ + π₯ = 2 (4,-1) Unit 6 Review 3. Write and solve a system of equations that represent this situation: Christine has $240 in her pocket. The cash is in tendollar and twenty-dollar bills. There are a total of 15 bills. How many of each type does she have? Unit 6 Review 3. Write and solve a system of equations that represent this situation: Christine has $240 in her pocket. The cash is in tendollar and twenty-dollar bills. There are a total of 15 bills. How many of each type does she have? 6 ten-dollar bills 9 twenty-dollar bills Unit 6 Review 4. Write and solve a system of equations that represent this situation: Wyattβs Custom Clocks sells handmade pedestal and wallhanging clocks. It costs the store $17 to buy the supplies to make a pedestal clock and $13 to buy the supplies to make a hanging clock. The store sells pedestal clocks for $56 and hanging clocks for $54. Last February Wyattβs Custom Clocks spent $379 on materials for pedestal and hanging clocks. They sold the finished products for a total of $1472. How many of each type of clock were sold in February? Unit 6 Review 4. Write and solve a system of equations that represent this situation: Wyattβs Custom Clocks sells handmade pedestal and wallhanging clocks. It costs the store $17 to buy the supplies to make a pedestal clock and $13 to buy the supplies to make a hanging clock. The store sells pedestal clocks for $56 and hanging clocks for $54. Last February Wyattβs Custom Clocks spent $379 on materials for pedestal and hanging clocks. They sold the finished products for a total of $1472. How many of each type of clock were sold in February? 7 pedestal clocks 20 hanging clocks Unit 6 Review 5. Solve. a. b. 1 π₯ 2 π¦= + 10 π₯ β 2π¦ = β8 π¦= β2 π₯ 3 +5 β2π₯ β 3π¦ = β15 Unit 6 Review 5. Solve. a. b. 1 π₯ 2 π¦= + 10 π₯ β 2π¦ = β8 π¦= β2 π₯ 3 No solution Infinitely many solutions +5 β2π₯ β 3π¦ = β15
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