Linear Equations

1. -K.THARUN SANGEETH
CSE-B
20G21A0572

2. WHAT IS A LINEAR SYSTEM?
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.

3. INTRODUCTION TO SYSTEMS OF LINEAR
EQUATIONS
A linear equation in n variables:
a0 + a1x1 + a2x2 + a3x3 + …... + anxn = b
a1,a2,a3,…,an, b: real number
a1: leading coefficient
x1: leading variable

4. A system of m linear equations in n variables:
a1x1 + a12x2 + a13x3 + · · · · + a1nxn = b1
a21x1 + a22x2 + a23x3 + · · · · + a2nxn = b2
a31x1 + a32x2 + a33x3 + · · · · + a3nxn = b3
. . . . .
. . . . .
. . . . .
am1x1 + am2x2 + am3x3 + · · · + amnxn = bm
Consistent:
A system of linear equations has at least one solution.
Inconsistent:
A system of linear equations has no solution.

5. 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.

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

7. 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.

8. Example-2 :
Solve the following
system of equations
by graphing.
– x + 3y = 6 and
3x – 9y = 9
First, graph – x + 3y = 6
Second, graph 3x-9y=9
The lines APPEAR to be parallel. Continued.

9. 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)
3
1 y = x + 2 (divide both sides by 3)
Second equation,
3x – 9y = 9
–9y = –3x + 9 (subtract 3x from both sides)
3
1 y = x – 1 (divide both sides by –9)
Both lines have a slope of , so they are parallel and do not intersect. Hence,
there is no solution to the system

10. 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
Rise is -10
because we
went down
Rise is 10
because we
went up
Run is 6
because we
went to the
right
Slope triangle between two
points. Notice that the slope
triangle can be drawn two
different ways.
The slope in this case is -10/-6 = 5/3
The slope in this case is 10/6 = 5/3

11. THE FORMULA IS USED WHEN YOU KNOW TWO
POINTS OF A LINE.
THEY LOOK LIKE A(X,Y) AND B(X,Y)
SLOPE=RISE/RUN = (Y2-Y1)/(X2-X1)
FORMULA FOR FINDING SLOPE

12. Find the slope of the line between the two points (-4, 8) and (10, -4)
If it helps label the points.
Then use the
Formula
(Y2-Y1)/(X2-X1) SUBSTITUTE INTO FORMULA [(-4)-(8)]/[(10)-(-4)]
Then Simplify
[(-4)-(8)]/[(10)-(-4)] = -12/14 = -6/7

13. Matrix equation
m x n matrix:
m rows
n columns
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 x n .
3) If m x n, then the matrix is called square of order n.
4) For a square matrix, the entries a11, a22, …. , ann are called the main
diagonal entries.

14. Matrix form: ax = b
Coefficient matrix:
= A

15. 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.

16. 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.

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

18. 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.
Example continued

19. THE ELIMINATION METHOD
THE CRYPTOGRAPHIC METHOD
Solving systems of linear equations
by addition

20. 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.
The Elimination Method

21. Solve the following system of equations using the
elimination method.
6x – 3y = –3 and 4x + 5y = –9
The Elimination Method
Example
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)
Continued.

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

23. 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).
The Elimination Method
Example continued
Continued.

24. 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).
The Elimination Method
Example continued

25. Use of matrix
1 2
0 3
To obtain the Hill cipher for the obtain text message
I AM HIDING
Example
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 Continued.

26. or equivalently
91 13 8 94 9 14 77
To cipher the pain text, we form the matrix product
1 2 9 11
0 3 1 = 3
Example Continued

27. 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

28. The entire cipher text message is
KC CX QL KP UU
which would usually be transmitted as a single string
Without spaces
KCCXQLKPUU