Linear eq

1. ELECTRICAL-1
MATHS GROUP
ID NUMBER:-31,32,33,34,35,36

2. Systems of Linear
Equations
How to: solve by graphing, substitution, linear
combinations, and special types of linear systems

3. What is a Linear System, Anyways?
A linear system includes two, or more,
equations, and each includes two or more
variables.
When two equations are used to model a
problem, it is called a linear system.

4. a linear equation in n variables:
a1,a2,a3,…,an, b: real number
a1: leading coefficient
x1: leading variable

5. a system of m linear equations in n variables:
5

a x  a x  a x   a x 
b
n n
11 1 12 2 13 3 1 1

a x  a x  a x   a x 
b
n n
21 1 22 2 23 3 2 2

a x  a x  a x   a x 
b
n n
31 1 32 2 33 3 3 3


a x  a x  a x   a x 
b
m 1 1 m 2 2 m 3 3
mn n m
 Consistent:
A system of linear equations has at least one solution.
 Inconsistent:
A system of linear equations has no solution.
Elementary Linear Algebra: Section 1.1, p.4

6. Finding a Solution by Graphing
• Since our chances of guessing the right
coordinates to try for a solution are not that high,
we’ll be more successful if we try a different
technique.
• Since a solution of a system of equations is a
solution common to both equations, it would also
be a point common to the graphs of both
equations.
• So to find the solution of a system of 2 linear
equations, graph the equations and see where
the lines intersect.

7. Finding a Solution by Graphing
Solve the following
system of equations
by graphing.
2x – y = 6 and
x + 3y = 10
x
y
First, graph 2x – y = 6.
(0, -6)
(3, 0)
(6, 6)
Second, graph x + 3y = 10
(-2, 4)
(1, 3)
(-5, 5)
The lines APPEAR to intersect at (4, 2).
(4, 2)
Example
Continued.

8. Finding a Solution by Graphing
Example continued
Although the solution to the system of equations
appears to be (4, 2), you still need to check the answer
by substituting x = 4 and y = 2 into the two equations.
First equation,
2(4) – 2 = 8 – 2 = 6 true
Second equation,
4 + 3(2) = 4 + 6 = 10 true
The point (4, 2) checks, so it is the solution of the
system.

9. Finding a Solution by Graphing
Example
Solve the following
system of equations
by graphing.
– x + 3y = 6 and
3x – 9y = 9
x
y
First, graph – x + 3y = 6
(-6, 0)
(0, 2)
(6, 4)
Second, graph
3x-9y=9
(0, -1)
(6, 1)
(3, 0)
The lines APPEAR to be parallel.
Continued.

10. Finding a Solution by Graphing
Example continued
Although the lines appear to be parallel, you still need to check
that they have the same slope. You can do this by solving for y.
First equation,
–x + 3y = 6
3y = x + 6 (add x to both sides)
1 y = x + 2 (divide both sides by 3)
3
Second equation,
3x – 9y = 9
–9y = –3x + 9 (subtract 3x from both sides)
1 y = x – 1 (divide both sides by –9)
3
Both lines have a slope of , so they are parallel and do not intersect. Hence,
there is no solution to the system

11. SLOPE
Slope is the ratio of the vertical rise to the horizontal
run between any two points on a line. Usually
referred to as the rise over run. Slope triangle between two
points. Notice that the slope
triangle can be drawn two
different ways.
Rise is -10
because we
went down
Run is -6
because we
went to the
left
5
3
10
6



Theslopein this case is
Rise is 10
because we
went up
Run is 6
because we
went to the
right
5
3
10
The slope in this case is 
6
Another way to
find slope

12. FORMULA FOR FINDING SLOPE
The formula is used when you know two
points of a line.
( , ) ( , ) 1 1 2 2 They look like A X Y and B X Y
Y 
Y
2 1
X X
2 1
RISE
RUN
SLOPE

 
EXAMPLE

13. Find the slope of the line between the two points (-4, 8) and (10, -4)
If it helps label the points. 1 X 1 Y
2 X 2 Y
Then use the
formula
Y 
Y
2 1
X 
X
2 1
(  4) 
(8)
(10)  ( 
4)
SUBSTITUTE INTO FORMULA
6
( 4) (8) 
7
12
14
 
(10) ( 4)



 
Then Simplify

14. Matrix equation
mn matrix:
14








n
11 12 13 1
n
n

a a a a

a a a a
21 22 23 2

a a a a
31 32 33 3


a a a a
 Notes:
(1) Every entry aij in a matrix is a number.
(2) A matrix with m rows and n columns is said to be of size mn .
(4) For a square matrix, the entries a11, a22, …, ann are called
the main diagonal entries.




m 1 m 2 m 3
mn
m rows
n columns
(3) If m  n, then the matrix is called square of order n.
Elementary Linear Algebra: Section 1.2, p.14

15. 












n
11 12 13 1
n
n

a a a a

a a a a
21 22 23 2
a a a a
31 32 33 3
a a a a
m m m mn
A



1 2 3











b
1
b
2
m b
b












x
x
1
2
n x
x

b Ax  Matrix form:
 Coefficient matrix:
A

a a a a
11 12 13 1

a a a a
21 22 23 2
a a a a
31 32 33 3
a a a a
n
n
n
m m m mn
















1 2 3

16. The Substitution Method
Another method (beside getting lucky with
trial and error or graphing the equations) that
can be used to solve systems of equations is
called the substitution method.
You solve one equation for one of the
variables, then substitute the new form of the
equation into the other equation for the
solved variable.

17. The Substitution Method
Solve the following system using the substitution method.
3x – y = 6 and – 4x + 2y = –8
Solving the first equation for y,
3x – y = 6
–y = –3x + 6 (subtract 3x from both sides)
y = 3x – 6 (multiply both sides by – 1)
Substitute this value for y in the second equation.
–4x + 2y = –8
–4x + 2(3x – 6) = –8 (replace y with result from first equation)
–4x + 6x – 12 = –8 (use the distributive property)
2x – 12 = –8 (simplify the left side)
2x = 4 (add 12 to both sides)
x = 2 (divide both sides by 2)
Example
Continued.

18. The Substitution Method
Example continued
Substitute x = 2 into the first equation solved for y.
y = 3x – 6 = 3(2) – 6 = 6 – 6 = 0
Our computations have produced the point (2, 0).
Check the point in the original equations.
First equation,
3x – y = 6
3(2) – 0 = 6 true
Second equation,
–4x + 2y = –8
–4(2) + 2(0) = –8 true
The solution of the system is (2, 0).

19. The Substitution Method
Solving a System of Linear Equations by the
Substitution Method
1) Solve one of the equations for a variable.
2) Substitute the expression from step 1 into
the other equation.
3) Solve the new equation.
4) Substitute the value found in step 3 into
either equation containing both variables.
5) Check the proposed solution in the original
equations.

20. The Substitution Method
Solve the following system of equations using the
substitution method.
y = 2x – 5 and 8x – 4y = 20
Since the first equation is already solved for y, substitute
this value into the second equation.
8x – 4y = 20
8x – 4(2x – 5) = 20 (replace y with result from first equation)
8x – 8x + 20 = 20 (use distributive property)
20 = 20 (simplify left side)
Example
Continued.

21. The Substitution Method
Example continued
When you get a result, like the one on the previous
slide, that is obviously true for any value of the
replacements for the variables, this indicates that
the two equations actually represent the same line.
There are an infinite number of solutions for this
system. Any solution of one equation would
automatically be a solution of the other equation.
This represents a consistent system and the linear
equations are dependent equations.

22. The Substitution Method
Solve the following system of equations using the substitution
method.
3x – y = 4 and 6x – 2y = 4
Solve the first equation for y.
3x – y = 4
–y = –3x + 4 (subtract 3x from both sides)
y = 3x – 4 (multiply both sides by –1)
Substitute this value for y into the second equation.
6x – 2y = 4
6x – 2(3x – 4) = 4 (replace y with the result from the first equation)
6x – 6x + 8 = 4 (use distributive property)
8 = 4 (simplify the left side)
Example
Continued.

23. The Substitution Method
Example continued
When you get a result, like the one on the previous
slide, that is never true for any value of the
replacements for the variables, this indicates that
the two equations actually are parallel and never
intersect.
There is no solution to this system.
This represents an inconsistent system, even though
the linear equations are independent.

24. Solving systems of linear equations
by addition

25. The Elimination Method
Another method that can be used to solve
systems of equations is called the addition or
elimination method.
You multiply both equations by numbers that
will allow you to combine the two equations
and eliminate one of the variables.

26. The Elimination Method
Solve the following system of equations using the elimination
method.
6x – 3y = –3 and 4x + 5y = –9
Multiply both sides of the first equation by 5 and the second
equation by 3.
First equation,
5(6x – 3y) = 5(–3)
30x – 15y = –15 (use the distributive property)
Second equation,
3(4x + 5y) = 3(–9)
12x + 15y = –27 (use the distributive property)
Example
Continued.

27. The Elimination Method
Example continued
Combine the two resulting equations (eliminating the
variable y).
30x – 15y = –15
12x + 15y = –27
42x = –42
x = –1 (divide both sides by 42)
Continued.

28. The Elimination Method
Example continued
Substitute the value for x into one of the original
equations.
6x – 3y = –3
6(–1) – 3y = –3 (replace the x value in the first equation)
–6 – 3y = –3 (simplify the left side)
–3y = –3 + 6 = 3 (add 6 to both sides and simplify)
y = –1 (divide both sides by –3)
Our computations have produced the point (–1, –1).
Continued.

29. The Elimination Method
Example continued
Check the point in the original equations.
First equation,
6x – 3y = –3
6(–1) – 3(–1) = –3 true
Second equation,
4x + 5y = –9
4(–1) + 5(–1) = –9 true
The solution of the system is (–1, –1).

30. The cryptographic method
Example
Use of matrix
1 2
0 3
To obtain the Hill cipher for the obain text message
I AM HIDING

31. The Cryptographic method
Example Continued
Solution
If we group the plaintext into pairs and add the
dummy letter G to fill out the last pair we obtain
IA MH ID IN GG

32. The cryptographic method
Example Continued
91 13 8 94 9 14 77
To cipher the paintext, we form the matrix product
1 2 9 11
=
0 3 1 3
or equivalently

33. The cryptographic method
Example Continued
Which from , yields ciphertext KC
To encipher the pair MH, we form the product
1 2 13 29
=
0 3 8 24

34. The cryptographic method
Example Continued
Whenever an integer greater than 25 occurs, it will be
by the remainder that results when this integer is
divided by 26
1 2 9 17
=
0 3 4 12
1 2 9 37 11
= or
0 3 14 42 16
1 2 7 21

35. The cryptographic method
Example Continued
The entire ciphertext message is
KC CX QL KP UU
which would usually be transmitted as a single string
Without spaces
KCCXQLKPUU