3 4 systems of linear equations 2

Systems of Linear Equations II
Frank Ma © 2011

There are two other methods to solve system of equations.

Substitution Method

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.

Substitution Method
2x + y = 7 E1
x + y = 5 E2
{
Solve
Example A.
by the substitution method.

Substitution Method
2x + y = 7 E1
x + y = 5 E2
{
Solve
Example A.
From E2, x + y = 5, we get x = 5 – y.

Substitution Method
2x + y = 7 E1
x + y = 5 E2
{
Solve
Example A.
From E2, x + y = 5, we get x = 5 – y. Then we replace
x by (5 – y) in E1

Substitution Method
2x + y = 7 E1
x + y = 5 E2
{
Solve
Example A.
x by (5 – y) in E1 and get
2(5 – y) + y = 7

Substitution Method
2x + y = 7 E1
x + y = 5 E2
{
Solve
Example A.
2(5 – y) + y = 7
10 – 2y + y = 7

Substitution Method
2x + y = 7 E1
x + y = 5 E2
{
Solve
Example A.
2(5 – y) + y = 7
10 – 2y + y = 7
10 – y = 7  3 = y

Substitution Method
2x + y = 7 E1
x + y = 5 E2
{
Solve
Example A.
2(5 – y) + y = 7
10 – 2y + y = 7
10 – y = 7  3 = y
Put 3 = y back to E2 to find x,

Substitution Method
2x + y = 7 E1
x + y = 5 E2
{
Solve
Example A.
2(5 – y) + y = 7
10 – 2y + y = 7
10 – y = 7  3 = y
Put 3 = y back to E2 to find x, we get
x + 3 = 5

Substitution Method
2x + y = 7 E1
x + y = 5 E2
{
Solve
Example A.
2(5 – y) + y = 7
10 – 2y + y = 7
10 – y = 7  3 = y
Put 3 = y back to E2 to find x, we get
x + 3 = 5  x = 2
Therefore the solution is (2, 3)

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

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
Example B.

2x – y = 7 E1
3x + 2y = 7 E2
{
Solve
Example B.
By inspection, we see that it's easy to solve for the y using E1.

2x – y = 7 E1
3x + 2y = 7 E2
{
Solve
Example B.
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.

2x – y = 7 E1
3x + 2y = 7 E2
{
Solve
Example B.
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

2x – y = 7 E1
3x + 2y = 7 E2
{
Solve
Example B.
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7

2x – y = 7 E1
3x + 2y = 7 E2
{
Solve
Example B.
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3

2x – y = 7 E1
3x + 2y = 7 E2
{
Solve
Example B.
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
To find y, use the substitution equation, set x = 3 in y = 2x – 7

2x – y = 7 E1
3x + 2y = 7 E2
{
Solve
Example B.
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
To find y, use the substitution equation, set x = 3 in y = 2x – 7 y = 2(3) – 7
y = -1

2x – y = 7 E1
3x + 2y = 7 E2
{
Solve
Example B.
3x + 2(2x – 7) = 7
3x + 4x – 14 = 7
7x = 21
x = 3
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).

Graphing Method

Graphing Method
Given a system of linear equations the graph of each equation is a straight line.

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.

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.

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.

Graphing Method
{
E1
2x + y = 7
x + y = 5
graphically.
Solve
Example C.
E2

Graphing Method
{
E1
2x + y = 7
x + y = 5
graphically.
Solve
Example C.
E2
Use intercept method.

Graphing Method
{
E1
2x + y = 7
x + y = 5
graphically.
Solve
Example C.
E2
2x + y = 7

Graphing Method
{
E1
2x + y = 7
x + y = 5
graphically.
Solve
Example C.
E2
E2
2x + y = 7

Graphing Method
{
E1
2x + y = 7
x + y = 5
graphically.
Solve
Example C.
E2
E2
2x + y = 7
(Check another point.)

Graphing Method
{
E1
2x + y = 7
x + y = 5
graphically.
Solve
Example C.
E2
E2
(0, 7)
2x + y = 7
(7/2, 0)
E1
E1

Graphing Method
{
E1
2x + y = 7
x + y = 5
graphically.
Solve
Example C.
E2
(0, 7)
x + y = 5
2x + y = 7
(7/2, 0)
E1

Graphing Method
{
E1
2x + y = 7
x + y = 5
graphically.
Solve
Example C.
E2
(0, 7)
2x + y = 7
x + y = 5
(7/2, 0)
E1

Graphing Method
{
E1
2x + y = 7
x + y = 5
graphically.
Solve
Example C.
E2
(0, 7)
(0, 5)
(5, 0)
2x + y = 7
x + y = 5
(7/2, 0)
E2
E1

Graphing Method
{
E1
2x + y = 7
x + y = 5
graphically.
Solve
Example C.
E2
(0, 7)
(0, 5)
The intersection
(2, 3) is the solution
(5, 0)
E1
2x + y = 7
x + y = 5
(7/2, 0)
E2
E1

Graphing of inconsistent systems are parallel lines which do not intersect, hence there is no solution.

{
E1
x + y = 7
x + y = 5
graphically.
Solve
Example D.
E2

{
E1
x + y = 7
x + y = 5
graphically.
Solve
Example D.
E2
x + y = 7

{
E1
x + y = 7
x + y = 5
graphically.
Solve
Example D.
E2
x + y = 7
(0, 7)
(7, 0)

{
E1
x + y = 7
x + y = 5
graphically.
Solve
Example D.
E2
x + y = 7
(0, 7)
x + y = 5
(7, 0)

{
E1
x + y = 7
x + y = 5
graphically.
Solve
Example D.
E2
x + y = 7
(0, 7)
(0, 5)
x + y = 5
(5, 0)
(7, 0)

{
E1
x + y = 7
x + y = 5
graphically.
Solve
Example D.
E2
x + y = 7
(0, 7)
No intersection.
No solution
(0, 5)
x + y = 5
(5, 0)
(7, 0)

Graphs of dependent systems are identical lines, hence every point on the line is a solution.

E1
x + y = 5
2x + 2y = 10
{
graphically.
Solve
Example E.
E2

E1
x + y = 5
2x + 2y = 10
{
graphically.
Solve
Example E.
E2
2x + 2y = 10 is the same as
x + y = 5

E1
x + y = 5
2x + 2y = 10
{
graphically.
Solve
Example E.
E2
x + y = 5
(0, 5)
(5, 0)

E1
x + y = 5
2x + 2y = 10
{
graphically.
Solve
Example E.
E2
x + y = 5
Every point is a solution, e.g.(0, 5), (2, 3), (5, 0)…
(0, 5)
(2, 3)
(5, 0)

3 4 systems of linear equations 2

by math123a on Jun 22, 2011

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3 4 systems of linear equations 2 — Presentation Transcript