1. 9.2 Systems of Linear Equations
in Two Variables
Luke 9:23 "And he said to all, “If anyone would come after
me, let him deny himself and take up his cross daily and follow
me."

2. Recall from your mathematical past ...

3. Recall from your mathematical past ...
Two coplanar lines can intersect in
0, 1 or infinitely many points

4. Recall from your mathematical past ...
Two coplanar lines can intersect in
0, 1 or infinitely many points
So ... a system of two linear equations:

5. Recall from your mathematical past ...
Two coplanar lines can intersect in
0, 1 or infinitely many points
So ... a system of two linear equations:
a) 1 intersection point --> exactly 1 solution

6. Recall from your mathematical past ...
Two coplanar lines can intersect in
0, 1 or infinitely many points
So ... a system of two linear equations:
a) 1 intersection point --> exactly 1 solution
b) 0 intersection points --> no solutions
parallel lines
inconsistent system

7. Recall from your mathematical past ...
Two coplanar lines can intersect in
0, 1 or infinitely many points
So ... a system of two linear equations:
a) 1 intersection point --> exactly 1 solution
b) 0 intersection points --> no solutions
parallel lines
inconsistent system
c) infinite number of intersection points
--> same line
infinite number of solutions
dependent system

8. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4

9. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4
⎧ 60x − 15y = 35
⎨
⎩ − 60x + 15y = 12

10. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4
⎧ 60x − 15y = 35
⎨
⎩ − 60x + 15y = 12
0 = 47

11. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4
⎧ 60x − 15y = 35
⎨
⎩ − 60x + 15y = 12
0 = 47
This is false ... therefore, INCONSISTENT
No Solutions

12. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4
⎧ 60x − 15y = 35
⎨
⎩ − 60x + 15y = 12
0 = 47
This is false ... therefore, INCONSISTENT
No Solutions
Let’s ‘Skin The Cat’ ...

13. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4

14. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4
Using Matrices:
⎡ 12 −3 ⎤ ⎡ 7 ⎤
A = ⎢ ⎥ B = ⎢ ⎥
⎣ −20 5 ⎦ ⎣ 4 ⎦

15. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4
Using Matrices:
⎡ 12 −3 ⎤ ⎡ 7 ⎤
A = ⎢ ⎥ B = ⎢ ⎥
⎣ −20 5 ⎦ ⎣ 4 ⎦
A B gives an error condition ... therefore No Solutions
−1

16. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4
Using Matrices:
⎡ 12 −3 ⎤ ⎡ 7 ⎤
A = ⎢ ⎥ B = ⎢ ⎥
⎣ −20 5 ⎦ ⎣ 4 ⎦
A B gives an error condition ... therefore No Solutions
−1
Let’s ‘Skin The Cat’ again...

17. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4

18. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4
Put them into Slope-Intercept Form:

19. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4
Put them into Slope-Intercept Form:
1. − 3y = −12x + 7
7
y = 4x −
3

20. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4
Put them into Slope-Intercept Form:
1. − 3y = −12x + 7 2. 5y = 20x + 4
7 4
y = 4x − y = 4x +
3 5

21. 1. Solve the system:
⎧ 12x − 3y = 7
⎨
⎩ − 20x + 5y = 4
Put them into Slope-Intercept Form:
1. − 3y = −12x + 7 2. 5y = 20x + 4
7 4
y = 4x − y = 4x +
3 5
Same slopes, different y-intercepts means
Parallel Lines ... and therefore ...
No Solutions

22. 2. Solve the system:
⎧ 2x − 3y = 12
⎨
⎩ 4x − 6y = 24

23. 2. Solve the system:
⎧ 2x − 3y = 12
⎨
⎩ 4x − 6y = 24
⎧ − 4x + 6y = −24
⎨
⎩ 4x − 6y = 24

24. 2. Solve the system:
⎧ 2x − 3y = 12
⎨
⎩ 4x − 6y = 24
⎧ − 4x + 6y = −24
⎨
⎩ 4x − 6y = 24
0=0

25. 2. Solve the system:
⎧ 2x − 3y = 12
⎨
⎩ 4x − 6y = 24
⎧ − 4x + 6y = −24
⎨
⎩ 4x − 6y = 24
0=0
This is true ... therefore, same lines ... DEPENDENT
Infinitely Many Solutions

26. 2. Solve the system:
⎧ 2x − 3y = 12
⎨
⎩ 4x − 6y = 24
⎧ − 4x + 6y = −24
⎨
⎩ 4x − 6y = 24
0=0
This is true ... therefore, same lines ... DEPENDENT
Infinitely Many Solutions
The fancy-schmancy way of writing the solutions is:
⎧ 2 ⎫
⎨( x, y ) : y = x − 4 ⎬
⎩ 3 ⎭

27. 2. Solve the system:
⎧ 2x − 3y = 12 HEY!!
⎨
⎩ 4x − 6y = 24 Try Matrices on
this one ...
⎧ − 4x + 6y = −24
⎨
⎩ 4x − 6y = 24
0=0
This is true ... therefore, same lines ... DEPENDENT
Infinitely Many Solutions
The fancy-schmancy way of writing the solutions is:
⎧ 2 ⎫
⎨( x, y ) : y = x − 4 ⎬
⎩ 3 ⎭

28. 2. Solve the system:
⎧ 2x − 3y = 12
⎨
⎩ 4x − 6y = 24
⎡ 2 −3 ⎤ ⎡ 12 ⎤
A = ⎢ ⎥ B = ⎢ ⎥
⎣ 4 −6 ⎦ ⎣ 24 ⎦

29. 2. Solve the system:
⎧ 2x − 3y = 12
⎨
⎩ 4x − 6y = 24
⎡ 2 −3 ⎤ ⎡ 12 ⎤
A = ⎢ ⎥ B = ⎢ ⎥
⎣ 4 −6 ⎦ ⎣ 24 ⎦
The same error condition appears.
So ... the Matrix Method has this limitation.
Know how to use your calculator!!

30. 3. Liz rows her boat upstream from one point on a
river to another point four miles away in three
hours. The return trip, traveling with the current,
takes two hours. How fast does Liz row relative to
the water, and at what speed is the current flowing?

31. 3. Liz rows her boat upstream from one point on a
river to another point four miles away in three
hours. The return trip, traveling with the current,
takes two hours. How fast does Liz row relative to
the water, and at what speed is the current flowing?
D = rt

32. 3. Liz rows her boat upstream from one point on a
river to another point four miles away in three
hours. The return trip, traveling with the current,
takes two hours. How fast does Liz row relative to
the water, and at what speed is the current flowing?
D = rt
c = rate of current
b = rate of boat

33. 3. Liz rows her boat upstream from one point on a
river to another point four miles away in three
hours. The return trip, traveling with the current,
takes two hours. How fast does Liz row relative to
the water, and at what speed is the current flowing?
D = rt
c = rate of current
b = rate of boat
Dup = ( b − c ) t1

34. 3. Liz rows her boat upstream from one point on a
river to another point four miles away in three
hours. The return trip, traveling with the current,
takes two hours. How fast does Liz row relative to
the water, and at what speed is the current flowing?
D = rt
c = rate of current
b = rate of boat
Dup = ( b − c ) t1
Ddown = ( b + c ) t 2

35. 3. Liz rows her boat upstream from one point on a
river to another point four miles away in three
hours. The return trip, traveling with the current,
takes two hours. How fast does Liz row relative to
the water, and at what speed is the current flowing?
D = rt
⎧ 4 = ( b − c ) 3
⎪
c = rate of current ⎨
⎪ 4 = ( b + c ) 2
⎩
b = rate of boat
Dup = ( b − c ) t1
Ddown = ( b + c ) t 2

36. 3. Liz rows her boat upstream from one point on a
river to another point four miles away in three
hours. The return trip, traveling with the current,
takes two hours. How fast does Liz row relative to
the water, and at what speed is the current flowing?
D = rt
⎧ 4 = ( b − c ) 3
⎪
c = rate of current ⎨
⎪ 4 = ( b + c ) 2
⎩
b = rate of boat
Dup = ( b − c ) t1 ⎧ 3b − 3c = 4
⎨
Ddown = ( b + c ) t 2 ⎩ 2b + 2c = 4

37. ⎧ 3b − 3c = 4
⎨
⎩ 2b + 2c = 4

38. ⎧ 3b − 3c = 4
⎨
⎩ 2b + 2c = 4
⎧ 6b − 6c = 8
⎨
⎩ 6b + 6c = 12

39. ⎧ 3b − 3c = 4
⎨
⎩ 2b + 2c = 4
⎧ 6b − 6c = 8
⎨
⎩ 6b + 6c = 12
12b = 20

40. ⎧ 3b − 3c = 4
⎨
⎩ 2b + 2c = 4
⎧ 6b − 6c = 8
⎨
⎩ 6b + 6c = 12
12b = 20
5
b=
3

41. ⎧ 3b − 3c = 4 ⎛ 5 ⎞
⎨ 1. 3 ⎜ ⎟ − 3c = 4
⎩ 2b + 2c = 4 ⎝ 3 ⎠
⎧ 6b − 6c = 8
⎨
⎩ 6b + 6c = 12
12b = 20
5
b=
3

42. ⎧ 3b − 3c = 4 ⎛ 5 ⎞
⎨ 1. 3 ⎜ ⎟ − 3c = 4
⎩ 2b + 2c = 4 ⎝ 3 ⎠
⎧ 6b − 6c = 8 5 − 3c = 4
⎨
⎩ 6b + 6c = 12
12b = 20
5
b=
3

43. ⎧ 3b − 3c = 4 ⎛ 5 ⎞
⎨ 1. 3 ⎜ ⎟ − 3c = 4
⎩ 2b + 2c = 4 ⎝ 3 ⎠
⎧ 6b − 6c = 8 5 − 3c = 4
⎨
⎩ 6b + 6c = 12 1 = 3c
12b = 20
5
b=
3

44. ⎧ 3b − 3c = 4 ⎛ 5 ⎞
⎨ 1. 3 ⎜ ⎟ − 3c = 4
⎩ 2b + 2c = 4 ⎝ 3 ⎠
⎧ 6b − 6c = 8 5 − 3c = 4
⎨
⎩ 6b + 6c = 12 1 = 3c
12b = 20 1
c=
3
5
b=
3

45. ⎧ 3b − 3c = 4 ⎛ 5 ⎞
⎨ 1. 3 ⎜ ⎟ − 3c = 4
⎩ 2b + 2c = 4 ⎝ 3 ⎠
⎧ 6b − 6c = 8 5 − 3c = 4
⎨
⎩ 6b + 6c = 12 1 = 3c
12b = 20 1
c=
3
5
b=
3
5
Liz rows at a rate of MPH (relative to the water)
3
1
and the current is flowing at a rate of MPH.
3

46. 4. A vintner fortifies wine that contains 10% alcohol
by adding a 60% alcohol solution to it. The resulting
mixture has an alcoholic strength of 12% and fills
1100 one-liter bottles. How many liters of wine and
of the alcohol solution does the vintner use?

47. 4. A vintner fortifies wine that contains 10% alcohol
by adding a 60% alcohol solution to it. The resulting
mixture has an alcoholic strength of 12% and fills
1100 one-liter bottles. How many liters of wine and
of the alcohol solution does the vintner use?
w: liters of wine used
A: liters of alcohol solution used

48. 4. A vintner fortifies wine that contains 10% alcohol
by adding a 60% alcohol solution to it. The resulting
mixture has an alcoholic strength of 12% and fills
1100 one-liter bottles. How many liters of wine and
of the alcohol solution does the vintner use?
w: liters of wine used
A: liters of alcohol solution used
⎧ w + A = 1100
⎨
⎩ .1w + .6A = .12 (1100 )

49. 4. A vintner fortifies wine that contains 10% alcohol
by adding a 60% alcohol solution to it. The resulting
mixture has an alcoholic strength of 12% and fills
1100 one-liter bottles. How many liters of wine and
of the alcohol solution does the vintner use?
w: liters of wine used
A: liters of alcohol solution used
⎧ w + A = 1100
⎨
⎩ .1w + .6A = .12 (1100 )
⎧ w + A = 1100
⎨
⎩ w + 6A = 1320

50. 4. A vintner fortifies wine that contains 10% alcohol
by adding a 60% alcohol solution to it. The resulting
mixture has an alcoholic strength of 12% and fills
1100 one-liter bottles. How many liters of wine and
of the alcohol solution does the vintner use?
w: liters of wine used
A: liters of alcohol solution used
⎧ − w − A = −1100
⎧ w + A = 1100 ⎨
⎨ ⎩ w + 6A = 1320
⎩ .1w + .6A = .12 (1100 )
⎧ w + A = 1100
⎨
⎩ w + 6A = 1320

51. 4. A vintner fortifies wine that contains 10% alcohol
by adding a 60% alcohol solution to it. The resulting
mixture has an alcoholic strength of 12% and fills
1100 one-liter bottles. How many liters of wine and
of the alcohol solution does the vintner use?
w: liters of wine used
A: liters of alcohol solution used
⎧ − w − A = −1100
⎧ w + A = 1100 ⎨
⎨ ⎩ w + 6A = 1320
⎩ .1w + .6A = .12 (1100 )
5A = 220
⎧ w + A = 1100
⎨
⎩ w + 6A = 1320

52. 4. A vintner fortifies wine that contains 10% alcohol
by adding a 60% alcohol solution to it. The resulting
mixture has an alcoholic strength of 12% and fills
1100 one-liter bottles. How many liters of wine and
of the alcohol solution does the vintner use?
w: liters of wine used
A: liters of alcohol solution used
⎧ − w − A = −1100
⎧ w + A = 1100 ⎨
⎨ ⎩ w + 6A = 1320
⎩ .1w + .6A = .12 (1100 )
5A = 220
⎧ w + A = 1100
⎨ A = 44
⎩ w + 6A = 1320

53. 4. A vintner fortifies wine that contains 10% alcohol
by adding a 60% alcohol solution to it. The resulting
mixture has an alcoholic strength of 12% and fills
1100 one-liter bottles. How many liters of wine and
of the alcohol solution does the vintner use?
w: liters of wine used
A: liters of alcohol solution used
⎧ − w − A = −1100
⎧ w + A = 1100 ⎨
⎨ ⎩ w + 6A = 1320
⎩ .1w + .6A = .12 (1100 )
5A = 220
⎧ w + A = 1100
⎨ A = 44 w = 1056
⎩ w + 6A = 1320

54. 4. A vintner fortifies wine that contains 10% alcohol
by adding a 60% alcohol solution to it. The resulting
mixture has an alcoholic strength of 12% and fills
1100 one-liter bottles. How many liters of wine and
of the alcohol solution does the vintner use?
w: liters of wine used
A: liters of alcohol solution used
⎧ − w − A = −1100
⎧ w + A = 1100 ⎨
⎨ ⎩ w + 6A = 1320
⎩ .1w + .6A = .12 (1100 )
5A = 220
⎧ w + A = 1100
⎨ A = 44 w = 1056
⎩ w + 6A = 1320
1056 L of wine & 44 L of alcohol solution

55. HW #3
Small deeds done are better than great deeds planned.
Peter Marshall