1. Solving Systems of Equations and Inequalities Section 3.1A-B Two variable linear equations Section 3.1C Matrices Resolution of linear systems Section 3.1D Three variable linear equations Section 3.1D Determinants & Inverses of Matrices Section 3.2A Solving Systems of Linear Inequalities Section 3.2B-D Linear Programming

2. Section 3.1 Two Variable Linear Equations  A set of two or more linear equations that each contain two variables.  These equations represent lines that will intersect, overlap, or be parallel to each other.

3.  When these lines intersect or overlap each other they are said to be consistent. This means they have at least one point of intersection. If there is only one point of intersection, the lines are independent. If there is more than one point of intersection, the lines are dependent.  When two lines do not intersect, they are said to be inconsistent because the lines are parallel. Section 3.1 Two Variable Linear Equations

4.  Check your understanding: Given the following two linear equations, determine the consistency & dependence of the system. 2x + y = 5 & 4x + 2y = 8 a.Consistent, independent b.Consistent, dependent c.Inconsistent

5. Section 3.1 Two Variable Linear Equations  Check your understanding: Given the following two linear equations, determine the consistency & dependence of the system. 2x + y = 5 & 4x + 2y = 8 a.Consistent, independent b.Consistent, dependent c.Inconsistent

6. Section 3.1 Two Variable Linear Equations  Check your understanding: Given the following two linear equations, determine the consistency & dependence of the system. 3x + 2y = 5 & 6x + 2y = 8 a.Consistent, independent b.Consistent, dependent c.Inconsistent

7. Section 3.1 Two Variable Linear Equations  Check your understanding: Given the following two linear equations, determine the consistency & dependence of the system. 3x + 2y = 5 & 6x + 2y = 8 a.Consistent, independent b.Consistent, dependent c.Inconsistent

8. Section 3.1 Graphing Linear Equations  Systems of two linear equations that intersect at one point (consistent & independent) can be graphed separately by finding ordered pairs of the solution set for each and then locate the coordinates for the point of intersection.

9. Section 3.1 Graphing Linear Equations In blue we have y = 3x - 2, and in red we have y = -x + 2. y = 3x - 2y = -x + 2 0 1 2 -5 -2 1 4 0 1 2 32103210

10. Section 3.1 Graphing Linear Equations In blue we have y = 2x + 1, and in red we have y = x + 3. y = 2x + 1y = x + 3 0 1 2 1 3 5 0 1 2 23452345

11. Section 3.1 Graphing Linear Equations In blue we have y = 2x + 1, and in red we have y = 2x + 3. y = 2x + 1y = 2x + 3 0 1 2 1 3 5 0 1 2 13571357

12. Homework  Section 3.1A pages 163-164 complete problems 1 – 17.

13. Section 3.1 Two Variable Linear Equations  Systems of two linear equations are usually solved using either the elimination method or the substitution method.  The elimination method uses a process of removing one of the two variables simultaneously from both equations through addition of equal but opposite coefficients. In blue we have y = 3x - 2, and in red we have y = -x + 2.

14. Section 3.1 Two Variable Linear Equations In blue we have y = 3x - 2, and in red we have y = -x + 2. Blue 3x – y = 2 Red x + y = 2 4x = 4 With the variable y having equal but opposite values we can add the two equations together and eliminate the y variable. We get 4x = 4. This allows x = 1, and by inserting x = 1 back into the equation we can see that y = 1.

15. Section 3.1 Two Variable Linear Equations  The substitution method uses a process of rewriting one of the two equations by isolating one of the variables and then substituting the equation into the other equation. In blue we have y = 3x - 2, and in red we have y = -x + 2. With both equations in the form of y equals, we can substitute the 2 nd equation into the first and we get: -x + 2 = 3x – 2. Solving for x we get -4x = -4 and x = 1.

16. Practice Substitution Method 2x + y = 10  y = -2x + 10 3x – 4y = -10  3x - 4(-2x + 10) = -10 3x + 8x – 40 = -10 11x – 40 = -10 11x = -10 + 40 = 30 x = 30 / 11 y = -2( 30 / 11 ) + 10  y = - 60 / 11 + 110 / 11 = 50 / 11 5x - y = 9  y = 5x - 9 2x + 4y = 42  2x + 4(5x - 9) = -10 2x + 20x – 45 = -10 22x – 45 = -10 22x = -10 + 45 = 35 x = 35 / 22 ; y = 5( 35 / 22 ) – 9 = 175 / 22 – 198 / 22 = - 23 / 22

17. Practice Elimination Method 2x + y = 10  4(2x + y = 10)  8x + 4y = 40 3x – 4y = -10 11x = 30 x = 30 / 11 2( 30 / 11 ) + y = 10  y = 110 / 11 - 60 / 11 = 50 / 11 ; y = 50 / 11 5x - y = 9  4(5x - y = 9)  20x - 4y = 36 2x + 4y = 42 22x = 78 x = 39 / 11 5( 39 / 11 ) - y = 9  y = 195 / 11 - 99 / 11 = 96 / 11 ; y = 96 / 11

18. Section 3.1 Two Variable Linear Equations  Check your understanding: Given the following two linear equations, use the substitution method to start the solution of the system. 5x + 3y = 22 6x + 2y = 20 a.Y = 10 – 6x b.Y = 10 – 3x c.Y = 10 + 3x

19. Section 3.1 Two Variable Linear Equations  Check your understanding: Given the following two linear equations, use the substitution method to start the solution of the system. 5x + 3y = 22 6x + 2y = 20 a.Y = 10 – 6x b.Y = 10 – 3x c.Y = 10 + 3x

20. Homework  Section 3.1B pages 173 – 174 problems 3 – 21 odd

21. Section 3.1 - Matrices  Matrix – A rectangle array of terms (elements) arranged in columns and rows. A matrix with m rows and n columns is called an m x n matrix, (read m by n matrix).  Matrices are also used to determine solutions for multiple variable linear equations. This technique can be used as an alternative to elimination or substitution methods.

22. Section 3.1 - Matrices a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 3 x 3 Matrix The first number indicates the row (horizontal) and the second number indicates the column number (vertical). Equal Matrices – Two matrices are equal if and only they have the same dimensions and are equal element by element. = YXYX 2x – 6 2y This expression states that Y = 2x – 6 and x = 2y. Using the substitution method, we see that Y = 2(2y) – 6 and so y = 2, x = 4.

23. Section 3.1 - Matrices Addition of Matrices – The sum of two m x n matrices is a m x n matrix in which the elements are the sum of the corresponding elements of the given matrices. = A + B -2 + (-6) 0 + 7 1 + (-1) 0 + 4 5 + (-3) -8 + 10 A = -2 0 1 0 5 -8 = -6 7 -1 4 -3 10 B Solve for A + B. = A + B -8 7 0 4 2 2

24. Section 3.1 - Matrices Subtraction of Matrices – The difference of two m x n matrices is equal to the sum A + (-B) where (-B) is the additive inverse of B. = A - B -2 - (-6) 0 - 7 1 - (-1) 0 - 4 5 - (-3) -8 - 10 A = -2 0 1 0 5 -8 = -6 7 -1 4 -3 10 B Solve for A - B. = A - B 4 -7 2 -4 8 -18

25. Section 3.1 - Matrices Scalar Product – The product of a scalar k and an m x n matrix A is an m x n matrix denoted by kA. Each element of kA equals k times the corresponding element of A. = kA 5(-2) 5(0 ) 5(1) 5(0) 5(5) 5(-8) A = -2 0 1 0 5 -8 = 5 k Solve for kA. = kA -10 0 5 0 25 -40

26. Section 3.1 Determinants and Inverses  A determinant is a square array of numbers (written within a pair of vertical lines) which represents a certain sum of products.  Calculating a 2 × 2 Determinant  In general, we find the value of a 2 × 2 determinant with elements a, b, c, d as follows: We multiply the diagonals (top left × bottom right first), (bottom left x top right) then subtract the first product minus the second. a b c d det = a b c d =ad - cb

27. Section 3.1 Determinants and Inverses a b c d det = a b c d =ad - cb Ex:

28. 3.1 Determinants & Inverses Solve for the inverse by crisscrossing the first and fourth terms and reversing the sign of the second and third term. Multiply this with the reciprocal of the determinant for the matrix ( ). Examples of inverses for a 2 x 2 matrix. This is the inverse of the given matrix. Determinant = {(4  -2) – (-16)} = -8-(-6) = -2  =

29. 3.1 Determinants & Inverses  For 2 x 2 matrices, the product of a determinant and the inverse of a matrix can be used to solve systems of two linear equations.

30. Determinants & Inverses If we have a two variable system, we can use the determinant and inverse of a matrix to find the solution to the system. We solved for the value of x and y using matrices and found that x = -1 and y = 1.

31. Determinants & Inverses If we have a two variable system, we can use the determinant and inverse of a matrix to find the solution to the system. We solved for the value of x and y using matrices and found that x = -1 and y = 1.

32. Homework  Section 3.1D pages 190-191 problems 2 - 6

33. Cramer’s Rule Cramer’s Rule begins with the solving of the determinant for the system followed by the determinants for each of the variables within the system. The determinant for each of the variables is calculated by first substituting the solution column values for the variable column values and then worked on as a 2 x 2 matrix.

34. Cramer’s Rule (continued)

35. Cramer’s Rule (conclusion)

36. Section 3.1 Three variable linear systems  Three linear equation systems are solved using the elimination and substitution methods.  The technique requires the isolation of one of the variables amongst the three equations.  This gets repeated for a 2 nd variable and the remaining variable is then determined.  Afterwards, the other variables are determined.

37. Section 3.1 Three variable linear systems  Sample X + 2y + 2z = 10 2x – y + 2z = 6 X – 3y + 2z = 1 Solve for x, y,& z. Isolate the z value first. Combine line 1 & 2 and then combine line 3 & 2. X + 2y + 2z = 10  X + 2y + 2z = 10 2x – y + 2z = 6 -2x +y – 2z = -6 -x + 3y = 4 X – 3y + 2z = 1  X – 3y + 2z = 1 2x – y + 2z = 6  -2x +y – 2z = -6 -x - 2y = -5

38. Section 3.1 Three variable linear systems  Sample X + 2y + 2z = 10 2x – y + 2z = 6 X – 3y + 2z = 1 Solve for x, y,& z. Isolate the x value next. Combine line 1 & 2 answer and the line 3 & 2 answer. X + 2y + 2z = 11  X + 2y + 2z = 11 2x – y + 2z = 6  -2x +y – 2z = -6 -x + 3y = 5 X – 3y + 2z = 1  X – 3y + 2z = 1 2x – y + 2z = 6  -2x +y – 2z = -6 -x - 2y = -5 -x + 3y = 5  -x + 3y = 5 -x -2y = -5  x + 2y = 5 5y = 10 Y = 2, x = 1 Solve for z by reinserting y & x values. 1 + 2(2) + 2z = 10 Z = 2.5