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Linear Equations

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Linear Equations
Linear Equations
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Linear Equations

  1. 1. -K.THARUN SANGEETH CSE-B 20G21A0572
  2. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 14. Matrix form: ax = b Coefficient matrix: = A
  15. 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. 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. 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. 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. 19. THE ELIMINATION METHOD THE CRYPTOGRAPHIC METHOD Solving systems of linear equations by addition
  20. 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. 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. 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. 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. 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. 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. 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. 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. 28. The entire cipher text message is KC CX QL KP UU which would usually be transmitted as a single string Without spaces KCCXQLKPUU

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