Unit 6 review

Pre-Algebra
Unit 6 Review
Unit 6 Review
1) Solve by graphing.
3π‘₯ + 4𝑦 = 16
1
𝑦 = π‘₯ βˆ’1
2
Unit 6 Review
1) Solve by graphing.
3π‘₯ + 4𝑦 = 16
1
𝑦 = π‘₯ βˆ’1
2
(4,1)
Unit 6 Review
2. Solve by substitution.
4π‘₯ + 6𝑦 = 10
2𝑦 + π‘₯ = 2
Unit 6 Review
2. Solve by substitution.
4π‘₯ + 6𝑦 = 10
2𝑦 + 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:
Wyatt’s 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 Wyatt’s 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:
Wyatt’s 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 Wyatt’s 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
𝑦=
+ 10
π‘₯ βˆ’ 2𝑦 = βˆ’8
𝑦=
βˆ’2
π‘₯
3
+5
βˆ’2π‘₯ βˆ’ 3𝑦 = βˆ’15
Unit 6 Review
5. Solve.
a.
b.
1
π‘₯
2
𝑦=
+ 10
π‘₯ βˆ’ 2𝑦 = βˆ’8
𝑦=
βˆ’2
π‘₯
3
No solution
Infinitely many solutions
+5
βˆ’2π‘₯ βˆ’ 3𝑦 = βˆ’15