Download Math Homework: Solving Systems of Equations and more Assignments Mathematics in PDF only on Docsity!
Harford Community College – MATH 018 Spring 2009
Homework #4: Chapter 4
Due: 3/5/
KEY
Directions: Solve each of the following problems on separate paper. Staple everything together
with your name on each sheet of paper and submit by the due date. Show work when necessary;
otherwise, the problem becomes all or nothing. Each problem will be scored on the 0, 1, 4,
4.5, 5 scale as detailed in the syllabus.
- Show whether the point (-2, -9) solves the following system of equations:
x y
x y
− x − y =
x − y =
(-2, -9) DOES NOT satisfy BOTH equations – it is NOT a solution
In 2) and 3), solve the system of equations by graphing :
x y
x y
x y
x y
In 4) and 5), solve the system of equations by substitution :
x y
x y
x y
x y x y
y
y
y
y
y
y y
y y
x = y +
x y
x y
x y
x y
x y
x y
x y
x y
{ ( x , y )| 3 x − 6 y = 9 }
In 8) – 12), solve the system of equations by any of the three methods:
x y
y x
x is already isolated so substitution is a good strategy to use
y
y
y
y y
y
y
x
x
x
y x
x y
x y
− + =−
− − =
6 12 30
6 4 16
32 4 10
23 2 8
x y
x y
x y
x y
8
7
16
14 16
16
16 14
=
=
=
y
y
y
4
13
4
13 4
4
4 13
4 7 7 20 7
4 7 20
210 2
7 22
10 2
7 2
10 8
7 2 4
2 4 10
=
=
=
=
=
x
x
x
x
x
x
x
x
x y
x y
x y
x y
x y
x y
x y
x y
x y
y
y
y
x
x
x
x
x
x
x y
In 13) – 15), solve – you MUST set up a system of equations to get credit :
- A comic book collector is willing to pay $50 for a mint-condition comic book and $5 for a
comic book in good-condition. If a seller brings the collector 80 comic books (each is either in mint or good condition) and receives $985 for the transaction, how many mint-condition and how many good-condition comic books were sold?
Let x = number of comics in mint condition y = number of comics in good condition
x y
x y y x
x
x
x
x
x
x x
x x
y = − x
13 comics in mint condition and 67 comics in good condition
- Solution A contains 40% alcohol, solution B contains 30% alcohol, and solution C contains
35% alcohol. How many liters of solution A and solution B are required to make 45 liters of solution C?
Let x = liters of solution A y = liters of solution B
^ (^ )
x y
x y y x
x
x
x
x
x
x x
x x
x x
x y
y = − x
22.5 liters of Solution A and 22.5 liters of Solution B are required for the mixture
