1. 9.3 Systems of Linear Equations
in Several Variables
Day 2
Matthew 18:21-22 "Then Peter came up and said to him, “Lord,
how often will my brother sin against me, and I forgive him?
As many as seven times?” Jesus said to him, “I do not say to
you seven times, but seventy times seven."

2. Inconsistent & Dependent Systems

3. Inconsistent & Dependent Systems
⎧ x + 7y − 7z = 3
⎪
Solve: ⎨ 3x + 14y − 16z = 17
⎪ 2x + 7y − 9z = 10
⎩

4. Inconsistent & Dependent Systems
⎧ x + 7y − 7z = 3
⎪
Solve: ⎨ 3x + 14y − 16z = 17
⎪ 2x + 7y − 9z = 10
⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3

5. Inconsistent & Dependent Systems
⎧ x + 7y − 7z = 3
⎪
Solve: ⎨ 3x + 14y − 16z = 17
⎪ 2x + 7y − 9z = 10
⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x + 7y − 7z = 3
⎪
⎨ − 7y + 5z = 8
⎪ − 7y + 5z = 4
⎩

6. Inconsistent & Dependent Systems
⎧ x + 7y − 7z = 3
⎪
Solve: ⎨ 3x + 14y − 16z = 17
⎪ 2x + 7y − 9z = 10
⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x + 7y − 7z = 3
⎪
⎨ − 7y + 5z = 8
⎪ − 7y + 5z = 4
⎩
−1⋅ Eq 2 + Eq 3 → Eq 3

7. Inconsistent & Dependent Systems
⎧ x + 7y − 7z = 3
⎪
Solve: ⎨ 3x + 14y − 16z = 17
⎪ 2x + 7y − 9z = 10
⎩
⎧ x + 7y − 7z = 3
−3⋅ Eq 1+ Eq 2 → Eq 2 ⎪
⎨ − 7y + 5z = 8
−2 ⋅ Eq 1+ Eq 3 → Eq 3 ⎪
⎩ 0 = −4
⎧ x + 7y − 7z = 3
⎪
⎨ − 7y + 5z = 8
⎪ − 7y + 5z = 4
⎩
−1⋅ Eq 2 + Eq 3 → Eq 3

8. Inconsistent & Dependent Systems
⎧ x + 7y − 7z = 3
⎪
Solve: ⎨ 3x + 14y − 16z = 17
⎪ 2x + 7y − 9z = 10
⎩
⎧ x + 7y − 7z = 3
−3⋅ Eq 1+ Eq 2 → Eq 2 ⎪
⎨ − 7y + 5z = 8
−2 ⋅ Eq 1+ Eq 3 → Eq 3 ⎪
⎩ 0 = −4
⎧ x + 7y − 7z = 3
⎪ false ... therefore
⎨ − 7y + 5z = 8 Inconsistent
⎪ − 7y + 5z = 4 (No Solutions)
⎩
−1⋅ Eq 2 + Eq 3 → Eq 3

9. Inconsistent & Dependent Systems
⎧ x − y + 5z = −2
⎪
Solve: ⎨ 3x + y + 3z = 2
⎪ 3x + 9y − 21z = 18
⎩

10. Inconsistent & Dependent Systems
⎧ x − y + 5z = −2
⎪
Solve: ⎨ 3x + y + 3z = 2
⎪ 3x + 9y − 21z = 18
⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
−3⋅ Eq 1+ Eq 3 → Eq 3

11. Inconsistent & Dependent Systems
⎧ x − y + 5z = −2
⎪
Solve: ⎨ 3x + y + 3z = 2
⎪ 3x + 9y − 21z = 18
⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
−3⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8
⎪ 12y − 36z = 24
⎩

12. Inconsistent & Dependent Systems
⎧ x − y + 5z = −2
⎪
Solve: ⎨ 3x + y + 3z = 2
⎪ 3x + 9y − 21z = 18
⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
−3⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8
⎪ 12y − 36z = 24
⎩
−3⋅ Eq 2 + Eq 3 → Eq 3

13. Inconsistent & Dependent Systems
⎧ x − y + 5z = −2
⎪
Solve: ⎨ 3x + y + 3z = 2 ⎧ x − y + 5z = −2
⎪ 3x + 9y − 21z = 18 ⎪
⎩ ⎨ 4y − 12z = 8
⎪
−3⋅ Eq 1+ Eq 2 → Eq 2 ⎩ 0=0
−3⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8
⎪ 12y − 36z = 24
⎩
−3⋅ Eq 2 + Eq 3 → Eq 3

14. Inconsistent & Dependent Systems
⎧ x − y + 5z = −2
⎪
Solve: ⎨ 3x + y + 3z = 2 ⎧ x − y + 5z = −2
⎪ 3x + 9y − 21z = 18 ⎪
⎩ ⎨ 4y − 12z = 8
⎪
−3⋅ Eq 1+ Eq 2 → Eq 2 ⎩ 0=0
−3⋅ Eq 1+ Eq 3 → Eq 3
true ... Dependent
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8
⎪ 12y − 36z = 24
⎩
−3⋅ Eq 2 + Eq 3 → Eq 3

15. Inconsistent & Dependent Systems
⎧ x − y + 5z = −2
⎪
Solve: ⎨ 3x + y + 3z = 2 ⎧ x − y + 5z = −2
⎪ 3x + 9y − 21z = 18 ⎪
⎩ ⎨ 4y − 12z = 8
⎪
−3⋅ Eq 1+ Eq 2 → Eq 2 ⎩ 0=0
−3⋅ Eq 1+ Eq 3 → Eq 3
true ... Dependent
⎧ x − y + 5z = −2 z can be any number;
⎪ infinite solutions!
⎨ 4y − 12z = 8
⎪ 12y − 36z = 24
⎩
−3⋅ Eq 2 + Eq 3 → Eq 3

16. Inconsistent & Dependent Systems
⎧ x − y + 5z = −2
⎪
Solve: ⎨ 3x + y + 3z = 2 ⎧ x − y + 5z = −2
⎪ 3x + 9y − 21z = 18 ⎪
⎩ ⎨ 4y − 12z = 8
⎪
−3⋅ Eq 1+ Eq 2 → Eq 2 ⎩ 0=0
−3⋅ Eq 1+ Eq 3 → Eq 3
true ... Dependent
⎧ x − y + 5z = −2 z can be any number;
⎪ infinite solutions!
⎨ 4y − 12z = 8
⎪ 12y − 36z = 24
⎩ Drop z from the system and
−3⋅ Eq 2 + Eq 3 → Eq 3
solve Eq. 2 for y in terms of z

17. Drop z from the system and solve Eq. 2 for y in terms of z
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8
⎪
⎩ 0=0

18. Drop z from the system and solve Eq. 2 for y in terms of z
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8
⎪
⎩ 0=0
4y = 12z + 8
y = 3z + 2

19. Drop z from the system and solve Eq. 2 for y in terms of z
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8
⎪
⎩ 0=0
4y = 12z + 8
y = 3z + 2
Use this in Eq. 1 and
solve for x in terms of z

20. Drop z from the system and solve Eq. 2 for y in terms of z
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8
⎪
⎩ 0=0
4y = 12z + 8
y = 3z + 2
Use this in Eq. 1 and
solve for x in terms of z
x − ( 3z + 2 ) + 5z = −2
x − 3z − 2 + 5z = −2
x = −2z

21. Drop z from the system and solve Eq. 2 for y in terms of z
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8 Since z can be any number ...
⎪ let z=k
⎩ 0=0
4y = 12z + 8
y = 3z + 2
Use this in Eq. 1 and
solve for x in terms of z
x − ( 3z + 2 ) + 5z = −2
x − 3z − 2 + 5z = −2
x = −2z

22. Drop z from the system and solve Eq. 2 for y in terms of z
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8 Since z can be any number ...
⎪ let z=k
⎩ 0=0
y = 3k + 2
4y = 12z + 8
y = 3z + 2
Use this in Eq. 1 and
solve for x in terms of z
x − ( 3z + 2 ) + 5z = −2
x − 3z − 2 + 5z = −2
x = −2z

23. Drop z from the system and solve Eq. 2 for y in terms of z
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8 Since z can be any number ...
⎪ let z=k
⎩ 0=0
y = 3k + 2
4y = 12z + 8
x = −2k
y = 3z + 2
Use this in Eq. 1 and
solve for x in terms of z
x − ( 3z + 2 ) + 5z = −2
x − 3z − 2 + 5z = −2
x = −2z

24. Drop z from the system and solve Eq. 2 for y in terms of z
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8 Since z can be any number ...
⎪ let z=k
⎩ 0=0
y = 3k + 2
4y = 12z + 8
x = −2k
y = 3z + 2
Then, our answer is:
Use this in Eq. 1 and
solve for x in terms of z
x − ( 3z + 2 ) + 5z = −2
x − 3z − 2 + 5z = −2
x = −2z

25. Drop z from the system and solve Eq. 2 for y in terms of z
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8 Since z can be any number ...
⎪ let z=k
⎩ 0=0
y = 3k + 2
4y = 12z + 8
x = −2k
y = 3z + 2
Then, our answer is:
Use this in Eq. 1 and
solve for x in terms of z ( −2k, 3k + 2, k )
x − ( 3z + 2 ) + 5z = −2
x − 3z − 2 + 5z = −2
x = −2z

26. Drop z from the system and solve Eq. 2 for y in terms of z
⎧ x − y + 5z = −2
⎪
⎨ 4y − 12z = 8 Since z can be any number ...
⎪ let z=k
⎩ 0=0
y = 3k + 2
4y = 12z + 8
x = −2k
y = 3z + 2
Then, our answer is:
Use this in Eq. 1 and
solve for x in terms of z ( −2k, 3k + 2, k )
x − ( 3z + 2 ) + 5z = −2 k is any constant ...
x − 3z − 2 + 5z = −2 called a parameter
t is often used as our
x = −2z parametric “variable”

27. Groups: Try this one:

28. Groups: Try this one:
⎧ x − 2y + z = 3 ⎧ x − 2y + z = 3
⎪
Solve: ⎨ 2x − 5y + 6z = 7 ⎪
⎨ − y + 4z = 1
⎪ 2x − 3y − 2z = 5 ⎪
⎩ ⎩ 0=0
−2 ⋅ Eq 1+ Eq 2 → Eq 2 z=t
−2 ⋅ Eq 1+ Eq 3 → Eq 3 y = 4t − 1
x − 2 ( 4t − 1) + t = 3
⎧ x − 2y + z = 3
⎪ x − 8t + 2 + t = 3
⎨ − y + 4z = 1
⎪ x = 7t + 1
⎩ y − 4z = −1
Eq 2 + Eq 3 → Eq 3
( 7t + 1, 4t − 1, t )

29. (Hand out this problem to students)
An invester has $100,000 to invest in three types of bonds:
Short-Term, Intermediate-Term, and Long-Term. How much
should be invested in each to satisfy these conditions:
a) short-term bonds pay 4% annually
b) intermediate-term bonds pay 6% annually
c) long-term bonds pay 8% annually
d) equal amounts are to be invested in Intermediate and
Long-Term bonds; and
e) The investor wants a total annual return of $6700 on
the investment plan.
(Recall ... I=Prt)

30. s = amount invested in Short-Term bonds
i = amount invested in Intermediate-Term bonds
l = amount invested in Long-Term bonds

31. s = amount invested in Short-Term bonds
i = amount invested in Intermediate-Term bonds
l = amount invested in Long-Term bonds
⎧ s + i + l = 100,000
⎪
⎨ .04s + .06i + .08l = 6700
⎪ i − l = 0
⎩

32. s = amount invested in Short-Term bonds
i = amount invested in Intermediate-Term bonds
l = amount invested in Long-Term bonds
⎧ s + i + l = 100,000
⎪
⎨ .04s + .06i + .08l = 6700
⎪ i − l = 0
⎩
using matrices ...

33. s = amount invested in Short-Term bonds
i = amount invested in Intermediate-Term bonds
l = amount invested in Long-Term bonds
⎧ s + i + l = 100,000
⎪
⎨ .04s + .06i + .08l = 6700
⎪ i − l = 0
⎩
using matrices ...
s = 10,000
i = 45,000
l = 45,000

34. HW # 6
Quiz Tomorrow!!
Opportunity is missed by most people because it is
dressed in overalls and looks like work.
Thomas Edison