Quiz #6b
Answers
July 15, 2014
Math 141
1. solve the system of equations by any method shown in class.
x + 3y + z = 10
7y − z + 2x = 21
41 = z + 13y + 4x





x
1
2
4
y
3
7
13
z
1
−1
1

10
21
41

 rref 


 → 


x
1
0
0
y
0
1
0
z
10
−3
0

7
1
0




Answer:
x = 7 − 10z
y = 1 + 3z
z = any number
2. Parametric solution:
x = −8 + 4z
y = 33 − 5z
z = any number
We can not buy a part of an animal. So all of the variables must be an integer. We also know
that all of the variables must be greater than or equal to zero.
x≥0
−8 + 4z ≥ 0
4z ≥ 8
z≥2
y≥0
33 − 5z ≥ 0
33 ≥ 5z
33/5 ≥ z
z ≤ 33
5 ≈ 6.6
z≥0
In addition we know that the variables can not be any larger than 25
x ≤ 25
−8 + 4z ≤ 24
4z ≤ 32
z≤8
y ≤ 25
33 − 5z ≤ 24
7 ≤ 5z
7/5 ≤ z
z ≥ 75 ≈ 1.4
z ≤ 25
Taken all together, we find that z = 2, 3, 4, 5, 6.