1. 3. Side = x + 5 P = 3(side) P = 3(x + 5) P = 3x + 15 4. Side = 2x – 1 P = 4(side) P = 4(2x – 1) P = 8x – 4 5. Side = 2x + 3 P = 5(side) P = 5(2x + 3) P = 10x + 15 6. Side = 2x – 3 P = 5(side) P = 5(2x – 3) P = 10x – 15 8. Width = x Length = 3x P = 72 P = 2(width) + 2(length) 72 = 2(x) + 2(3x) 72 = 2x + 6x 72 = 8x 9 = x Width = 9, Length = 27 10.1 st Side = x 2 nd Side = x – 2 3 rd Side = x + 12 P = 73 P = Side1 + Side2 + Side3 73 = x + x – 2 + x + 12 73 = 3x +10 63 = 3x 21 = x Side 1 = 21, Side 2 = 19, Side 3 = 33 Homework Answers

2. 12.Width = x Length = x + 5 P = 66 P = 2(width) + 2(length) 66 = 2(x) + 2(x + 5) 66 = 2x + 2x + 10 66 = 4x + 10 56 = 4x 14 = x Width = 14, Length = 19 13. Width = x – 3 Length = x P = 130 P = 2(width) + 2(length) 130 = 2(x – 3) + 2(x) 130 = 2x – 6 + 2x 130 = 4x – 6 136 = 4x 34 = x Width = 31, Length = 34 Homework Answers

3. Lesson 25 Word Problems – Coin/Money

4. Coin/Money Word Problems Coin/Money problems are similar to Perimeter problems. Use information about one coin to create an expression for another coin. Multiply each coin expression by their worth.

5. Example of Coin problem In a collection of coins, there are 7 less dimes than quarters. If the total of the coin collection is $3.85, how many dimes and quarters are there? Step 1. Create expressions for each coin. Dimes = x – 7 Quarters = x

6. Step 2. Multiply each expression by their value and create and equation. Dimes =.10(x – 7) Quarters =.25x $3.85 =.10(x – 7) +.25x

7. Step 3. MOVE THE DECIMAL and solve the equation. 3.85 =.10(x – 7) +.25x 385 = 10(x – 7) + 25x 385 = 10x – 70 + 25x 385 = 35x – 70 +70 +70 455 = 35x 35 = 35 13 = x

8. Step 4. Go back to the expressions and find out how many coins there are. Dimes = x – 7 (13) – 7 = 6 Quarters = x 13 So there are 6 Dimes and 13 Quarters

9. Another Problem A collection of coins is worth $12.85. If the number of nickels is 6 less than twice the number of dimes and the number of quarters are 6 more than twice the number of dimes, how many nickels are there?