C = 5d + 2 2c + d = 4 Do Now. Homework Solutions 4)5x – 3y = – 4 15x – 9y = – 12 3x + 2y = 9...

c = 5d + 2

2c + d = 4

Do Now

Homework Solutions4) 5x – 3y = – 4 15x – 9y = – 12

3x + 2y = 9 – 15x + 10y = 45 – 19y = – 57

y = 33x + 2y = 93x + 2(3) = 93x + 6 = 93x = 3 x = 1

Solution: (1,3)

Homework Solutions

7) 3c – 3d = – 12 3c – 3d = – 12

c + 2d = 2 – 3c + 6d = 6

– 9d = – 18

d = 2

c + 2d = 2

c + 2(2) = 2

c + 4 = 2

c = – 2

Solution: (– 2,2)

Homework Solutions9) 4x – 3y = 5

y = 2x + 1

4x – 3(2x + 1) = 54x – 6x – 3 = 5 – 2x – 3 = 5 – 2x = 8

x = – 4 y = 2x + 1y = 2(– 4) + 1y = – 7

Solution: (– 4, – 7)

Homework Solutions11) 5x + 2y = 8

y = 1 – x

5x + 2(1 – x) = 85x + 2 – 2x = 8 3x + 2 = 8 3x = 6

x = 2 y = 1 – xy = 1 – (2)y = – 1

Solution: (2, – 1)

Word Problems

For Systems of Equations

CDs The play

Pizza

The sum of two numbers is 48 and their difference is 24. What are the numbers?

Three times one number added to another number is 18. Twice the first number minus the other number is 12. Find the numbers.

Three shirts and two neckties cost $69. At the same price, two shirts and three neckties cost $61. What is the cost of one shirt and one necktie?

Nicole bought 3 new compact discs and 3 used compactdiscs for $69.00. At the same prices, Jon bought 3 newcompact discs and 5 used compact discs for $85.00. Find the cost of buying a new and a used compact disc.

Let n = the cost of a new discLet u = the cost of an used disc

Set up two equations: 3n + 3u = $69.00 ** 3n + 5u = $85.00

Answer: New disc is $15 and Old disc is $8

Define the variables:

Brandon and Rachel had lunch at the mall.Brandon ordered three slices of pizza and two colas. Rachel ordered two slices of pizza and three colas. Brandon’s bill was$6.00 and Rachel’s bill was $5.25. Whatwas the price of one slice of pizza? What was the price of one cola?

Let p = the price of one slice of pizzaLet c = the price of one cola

3p + 2c = $6.002p + 3c = $5.252(3p + 2c = $6.00) 3(2p + 3c = $5.25)

Answer: 1 slice = 1.50, 1 cola = .75

The playThe total attendance at a schoolplay was 1250. The cost of the tickets were $6.00 for students and $7.50 for adults. The school drama club had a revenue total of $8362.50. How many of each ticket was sold?

Let a = the number of adultsLet s = the number of students

a + s = 1250 $7.50 a + $6.00s = $8362.50

$7.50(a + s = 1250 )

Answer: Adults = 575, Students = 675