Download Notes/Math/9R Systems Word Problems 1

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c = 5d + 2
2c + d = 4
Homework Solutions
4) 5x – 3y = – 4
3x + 2y = 9
3x + 2y = 9
3x + 2(3) = 9
3x + 6 = 9
3x = 3
x=1
Solution:
15x – 9y = – 12
– 15x + 10y = 45
– 19y = – 57
y = 3
(1,3)
Homework Solutions
7) 3c – 3d = – 12
c + 2d = 2
–
3c – 3d
3c + 6d
– 9d
d
c + 2d = 2
c + 2(2) = 2
c+4=2
c = –2
Solution:
(– 2,2)
= – 12
=
6
= – 18
= 2
Homework Solutions
9) 4x – 3y = 5
y = 2x + 1
4x – 3(2x + 1) = 5
4x – 6x – 3 = 5
– 2x – 3 = 5
– 2x = 8
x = –4
Solution:
(– 4, – 7)
y = 2x + 1
y = 2(– 4) + 1
y = –7
Homework Solutions
11) 5x + 2y = 8
y = 1 – x
5x + 2(1 – x) = 8
5x + 2 – 2x = 8
3x + 2 = 8
3x = 6
x = 2
Solution:
(2, – 1)
y = 1 – x
y = 1 – (2)
y = –1
Word Problems
For Systems of Equations
Pizza
CDs
The play
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 compact
discs for $69.00. At the same prices, Jon bought 3 new
compact discs and 5 used compact discs for $85.00. Find
the cost of buying a new and a used compact disc.
Define the variables:
Set up two equations:
Let n = the cost of a new disc
Let u = the cost of an used disc
3n + 3u = $69.00 **
3n + 5u = $85.00
Answer: New disc is $15 and Old disc is $8
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. What
was the price of one slice of pizza? What
was the price of one cola?
Let p = the price of one slice of pizza
Let c = the price of one cola
3p
+ 2c
= $6.00
2(3p
+ 2c
= $6.00)
2p
+ 3c
= $5.25
3(2p
+ 3c
= $5.25)
Answer: 1 slice = 1.50,
1 cola = .75
The total attendance at a school
play 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?
The play
Let a = the number of adults
Let s = the number of students
$7.50(a
s=
1250 )
a
++ s=
1250
$7.50 a + $6.00s = $8362.50
Answer: Adults = 575,
Students = 675