The document provides an example of solving a system of linear equations using the substitution method. It begins with the system 2x + y = 7 and x + y = 5. It solves the second equation for x in terms of y, getting x = 5 - y. This expression for x is then substituted into the first equation, giving 10 - 2y + y = 7, which can be solved to find the value of y, and then substituted back into the original equation to find the value of x. The solution is presented as (2, 3). The document then provides two additional examples demonstrating how to set up and solve systems of equations using the substitution method.

1. Systems of Linear Equations II

2. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?

3. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.

4. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.

5. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
Exchanging the 2 coupons for pizzas and donuts,
we would have
5P + 3D + 2C

6. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
Exchanging the 2 coupons for pizzas and donuts,
we would have
5P + 3D + 2C
= 5P + 3D + 2(5P + 3D)

7. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
Exchanging the 2 coupons for pizzas and donuts,
we would have
5P + 3D + 2C
= 5P + 3D + 2(5P + 3D)
= 5P + 3D + 10P + 6D

8. Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
Exchanging the 2 coupons for pizzas and donuts,
we would have
5P + 3D + 2C
= 5P + 3D + 2(5P + 3D)
= 5P + 3D + 10P + 6D
= 15P + 9D
or 15 slices of pizzas and 9 donuts.

9. In math, the phrase "substitute (the expression) back into ...“
means to do the exchange, using the given coupon–expression,
in the targeted equations, or expressions mentioned.
Systems of Linear Equations II
Example A. At Pizza Grande, one coupon C may be exchanged
for three slices of pizza (P) and five donuts (D).
We have 5 slices of pizzas, 3 donuts and 2 coupons.
What do we have after exchanging the 2 coupons?
The coupon value of C may be recorded as C = 5P + 3D.
5 slices of pizzas, 3 donuts and 2 coupons is 5P + 3D+ 2C.
Exchanging the 2 coupons for pizzas and donuts,
we would have
5P + 3D + 2C
= 5P + 3D + 2(5P + 3D)
= 5P + 3D + 10P + 6D
= 15P + 9D
or 15 slices of pizzas and 9 donuts.

10. There are two other methods to solve system of equations.
Systems of Linear Equations II

11. There are two other methods to solve system of equations.
Substitution Method
Systems of Linear Equations II

12. There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
Systems of Linear Equations II

13. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.

14. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.

15. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y

16. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Then we replace x by (5 – y) in E1 and get
2(5 – y) + y = 7
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y

17. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Then we replace x by (5 – y) in E1 and get
2(5 – y) + y = 7
10 – 2y + y = 7
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y

18. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Then we replace x by (5 – y) in E1 and get
2(5 – y) + y = 7
10 – 2y + y = 7
10 – y = 7  3 = y
Systems of Linear Equations II
Example B. {
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y

19. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Then we replace x by (5 – y) in E1 and get
2(5 – y) + y = 7
10 – 2y + y = 7
10 – y = 7  3 = y
Systems of Linear Equations II
Example B. {
Put 3 = y back to E2 to find x, we get
x + 3 = 5
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y

20. 2x + y = 7 E1
x + y = 5 E2
Solve
by the substitution method.
From E2, x + y = 5, we get x = 5 – y.
Then we replace x by (5 – y) in E1 and get
2(5 – y) + y = 7
10 – 2y + y = 7
10 – y = 7  3 = y
Systems of Linear Equations II
Example B. {
Put 3 = y back to E2 to find x, we get
x + 3 = 5  x = 2
Therefore the solution is (2, 3)
There are two other methods to solve system of equations.
Substitution Method
In substitution method, we solve for one of the variables in
terms of the other, then substitute the result into the other
equation.
This is the coupon expression
so we may exchange x = 5 – y

21. We use the substitution method when it's easy to solve for
one of the variable in terms of the other.
Systems of Linear Equations II

22. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
Systems of Linear Equations II

23. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
Systems of Linear Equations II
Example C. {

24. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
Systems of Linear Equations II
Example C. {

25. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y.
Systems of Linear Equations II
Example C. {

26. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
Systems of Linear Equations II
Example C. {

27. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
Systems of Linear Equations II
Example C. {

28. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
Systems of Linear Equations II
Example C. {

29. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
Systems of Linear Equations II
Example C. {
To find y, use the substitution equation, set x = 3 in y = 2x – 7

30. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
Systems of Linear Equations II
Example C. {
To find y, use the substitution equation, set x = 3 in y = 2x – 7
y = 2(3) – 7
y = -1

31. We use the substitution method when it's easy to solve for
one of the variable in terms of the other. Specifically, it is
easy to solve for a variable when an equation contains an
single x or single y.
2x – y = 7 E1
3x + 2y = 7 E2
Solve
by the substitution method.
By inspection, we see that it's easy to solve for the y using E1.
From E1, 2x – y = 7, we get 2x – 7= y. Substitute y by (2x – 7)
in E2 and get
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
Systems of Linear Equations II
Example C. {
To find y, use the substitution equation, set x = 3 in y = 2x – 7
y = 2(3) – 7
y = -1 Hence the solution is (3, -1).

32. Graphing Method
Systems of Linear Equations II

33. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line.
Systems of Linear Equations II

34. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines.
Systems of Linear Equations II

35. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically.
Systems of Linear Equations II

36. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II

37. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.

38. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.

39. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0
0
2x + y = 7

40. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
0
2x + y = 7

41. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7
E2

42. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7
E2
(Check another point.)

43. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7
E1
E2
(0, 7)
(7/2, 0)
E1(Check another point.)

44. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x + y = 5
x y
0 7
7/2 0
2x + y = 7
x y
0
0
(0, 7)
(7/2, 0)
E1(Check another point.)

45. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7 x + y = 5
x y
0 5
5 0
(0, 7)
(7/2, 0)
E1(Check another point.)

46. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7
(0, 7)
(7/2, 0)
(0, 5)
(5, 0)
E2
x + y = 5
x y
0 5
5 0
E1(Check another point.)

47. Graphing Method
Given a system of linear equations the graph of each equation
is a straight line. The solution of the system is the intersection
point (x, y) of the these lines. Thus we may find the solution by
graphing the lines and locate the point of intersection
graphically. In general, we don't not use the graphing method
because it is not easy to do it accurately.
Systems of Linear Equations II
2x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example D.
Use intercept method.
x y
0 7
7/2 0
2x + y = 7
(0, 7)
(7/2, 0)
(0, 5)
(5, 0)
The intersection
(2, 3) is the solution
E1
E2
x + y = 5
x y
0 5
5 0
E1(Check another point.)

48. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
Systems of Linear Equations II

49. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.

50. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
x + y = 7
x y
0 7
7 0
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.

51. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
x + y = 7
x y
0 7
7 0
(0, 7)
(7, 0)
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.

52. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
x + y = 7
x y
0 7
7 0
x y
0 5
5 0
(0, 7)
(7, 0)
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.
x + y = 5

53. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
x + y = 7
x y
0 7
7 0
x y
0 5
5 0
(0, 7)
(7, 0)
(0, 5)
(5, 0)
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.
x + y = 5

54. Graphing of inconsistent systems are parallel lines which do
not intersect, hence there is no solution.
x + y = 7
x y
0 7
7 0
x y
0 5
5 0
(0, 7)
(7, 0)
(0, 5)
(5, 0)
No intersection.
No solution
Systems of Linear Equations II
x + y = 7
x + y = 5
Solve graphically.{ E1
E2
Example E.
x + y = 5

55. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
Systems of Linear Equations II

56. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
Systems of Linear Equations II
x + y = 5
2x + 2y = 10
Solve graphically.{
E1
E2Example F.

57. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
2x + 2y = 10 is the same as
x + y = 5
Systems of Linear Equations II
x + y = 5
2x + 2y = 10
Solve graphically.{
E1
E2Example F.

58. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
2x + 2y = 10 is the same as
x + y = 5
x y
0 5
5 0
Systems of Linear Equations II
x + y = 5
2x + 2y = 10
Solve graphically.{
E1
E2Example F.

59. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
2x + 2y = 10 is the same as
x + y = 5
x y
0 5
5 0
(5, 0)
(0, 5)
Systems of Linear Equations II
x + y = 5
2x + 2y = 10
Solve graphically.{
E1
E2Example F.

60. Graphs of dependent systems are identical lines, hence every
point on the line is a solution.
2x + 2y = 10 is the same as
x + y = 5
x y
0 5
5 0
(5, 0)
(0, 5)
Every point is a solution,
e.g.(0, 5), (2, 3), (5, 0)…
(2, 3)
Systems of Linear Equations II
x + y = 5
2x + 2y = 10
Solve graphically.{
E1
E2Example F.

61. Systems of Linear Equations II
4. {–x + 2y = –12
y = 4 – 2x
Exercise. Solve by the substitution method.
1. {y = 3 – x
2x + y = 4
2. 3. {x = 3 – y
2x – y = 6
{x + y = 3
2x + 6 = y
5. {3x + 4y = 3
x = 6 + 2y
6. { x = 3 – 3y
2x – 9y = –4
10. Graph the inconsistent system
{ x + 3y = 4
2x + 6y = 8
{2x – y = 2
8x – 4y = 6
Problem 7, 8, 9: Solve problem 1, 2, and 3 by graphing.
11. Graph the dependent system