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Review 2.1-2.3

Review 2.1-2.3. (1,4) 1-3(4)= -5 1-12= -5 -11 = -5 *doesn ’ t work in the 1 st eqn, no need to check the 2 nd . Not a solution. (-5,0) -5-3(0)= -5 -5 = -5 -2(-5)+3(0)=10 10=10 Solution. Ex: Check whether the ordered pairs are solutions of the system. x-3y= -5 -2x+3y=10.

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Review 2.1-2.3

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  1. Review 2.1-2.3

  2. (1,4) 1-3(4)= -5 1-12= -5 -11 = -5 *doesn’t work in the 1st eqn, no need to check the 2nd. Not a solution. (-5,0) -5-3(0)= -5 -5 = -5 -2(-5)+3(0)=10 10=10 Solution Ex: Check whether the ordered pairs are solutions of the system.x-3y= -5-2x+3y=10

  3. Solving a System Graphically • Graph each equation on the same coordinate plane. (USE GRAPH PAPER!!!) • If the lines intersect: The point (ordered pair) where the lines intersect is the solution. • If the lines do not intersect: • They are the same line – infinitely many solutions (they have every point in common). • They are parallel lines – no solution (they share no common points).

  4. Ex: Solve the system graphically.2x-2y= -82x+2y=4 (-1,3)

  5. Ex: Solve the system graphically.2x+4y=12x+2y=6 1st eqn: y = -½x + 3 2ND eqn: y = -½x + 3 What does this mean? the 2 equations are for the same line! • Infinite many solutions

  6. Ex: Solve graphically: x-y=5 2x-2y=9 1st eqn: y = x – 5 2nd eqn: y = x – 9/2 What do you notice about the lines? They are parallel! No solution!

  7. Solving Systems of Equations using Substitution Steps: 1. Solve one equation for one variable (y= ; x= ; a=) 2. Substitute the expression from step one into the other equation. Then solve. 3. Substitute back into Step 1 and solve for the other variable. 4. Check the solution in both equations of the system.

  8. 1) Solve the system using substitution x + y = 5 y = 3 + x Step 1: Solve an equation for one variable. The second equation is already solved for y! Step 2: Substitute x + y = 5x + (3 + x) = 5 2x + 3 = 5 2x = 2 x = 1

  9. 1) Solve the system using substitution x + y = 5 y = 3 + x y = 3 + x Y = 3 + (1) y = 4 Step 3: Plug back in to find the other variable. (1, 4) (1) + (4) = 5 (4) = 3 + (1) Step 4: Check your solution. The solution is (1, 4). What do you think the answer would be if you graphed the two equations?

  10. Which answer checks correctly? • (2, 2) • (5, 3) • (3, 5) • (3, -5) 3x – y = 4 x = 4y - 17

  11. 2) Solve the system using substitution 3y + x = 7 4x – 2y = 0 It is easiest to solve the first equation for x. 3y + x = 7 -3y -3y x = -3y + 7 Step 1: Solve an equation for one variable. Step 2: Substitute 4x – 2y = 0 4(-3y + 7) – 2y = 0

  12. 2) Solve the system using substitution 3y + x = 7 4x – 2y = 0 -12y + 28 – 2y = 0 -14y + 28 = 0 -14y = -28 y = 2 x = -3y + 7 x = -3(2) + 7 x = -6 + 7 x = 1 Step 3: Plug back in to find the other variable.

  13. 2) Solve the system using substitution 3y + x = 7 4x – 2y = 0 Step 4: Check your solution. (1, 2) 3(2) + (1) = 7 4(1) – 2(2) = 0 When is solving systems by substitution easier to do than graphing? When only one of the equations has a variable already isolated (like in example #1).

  14. 3) Solve the system using substitution x = 3 – y x + y = 7 Step 1: Solve an equation for one variable. The first equation is already solved for x! Step 2: Substitute x + y = 7 (3 – y) + y = 7 3 = 7 The variables were eliminated!! This is a special case. Does 3 = 7? FALSE! When the result is FALSE, the answer is NO SOLUTIONS.

  15. 3) Solve the system using substitution 2x + y = 4 4x + 2y = 8 Step 1: Solve an equation for one variable. The first equation is easiest to solved for y! y = -2x + 4 4x + 2y = 8 4x + 2(-2x + 4) = 8 Step 2: Substitute 4x – 4x + 8 = 8 8 = 8 This is also a special case. Does 8 = 8? TRUE! When the result is TRUE, the answer is INFINITELY MANY SOLUTIONS.

  16. What does it mean if the result is “TRUE”? • The lines intersect • The lines are parallel • The lines are coinciding • The lines reciprocate • I can spell my name

  17. Solving Systems of Equations using Elimination Steps: 1. Place both equations in Standard Form Ax + By = C. 2. Determine which variable to eliminate with Addition or Subtraction. 3. Solve for the variable left. 4. Go back and use the found variable in step 3 to find second variable. 5. Check the solution!!!!

  18. 1) Solve the system using elimination. 2x + 2y = 6 3x – y = 5 Step 1: Put the equations in Standard Form. They already are! None of the coefficients are the same! Find the least common multiple of each variable. LCM = 6x, LCM = 2y Which is easier to obtain? 2y(you only have to multiplythe bottom equation by 2) Step 2: Determine which variable to eliminate.

  19. 1) Solve the system using elimination. 2x + 2y = 6 3x – y = 5 Multiply the bottom equation by 2 2x + 2y = 6 (2)(3x – y = 5) 8x = 16 x = 2 2x + 2y = 6 (+) 6x – 2y = 10 Step 3: Multiply the equations and solve. 2(2) + 2y = 6 4 + 2y = 6 2y = 2 y = 1 Step 4: Plug back in to find the other variable.

  20. 1) Solve the system using elimination. 2x + 2y = 6 3x – y = 5 (2, 1) 2(2) + 2(1) = 6 3(2) - (1) = 5 Step 5: Check your solution. Solving with multiplication adds one more step to the elimination process.

  21. 2) Solve the system using elimination. x + 4y = 7 4x – 3y = 9 Step 1: Put the equations in Standard Form. They already are! Find the least common multiple of each variable. LCM = 4x, LCM = 12y Which is easier to obtain? 4x(you only have to multiplythe top equation by -4 to make them inverses) Step 2: Determine which variable to eliminate.

  22. 2) Solve the system using elimination. x + 4y = 7 4x – 3y = 9 Multiply the top equation by -4 (-4)(x + 4y = 7) 4x – 3y = 9) y = 1 -4x – 16y = -28 (+) 4x – 3y = 9 Step 3: Multiply the equations and solve. -19y = -19 x + 4(1) = 7 x + 4 = 7 x = 3 Step 4: Plug back in to find the other variable.

  23. 2) Solve the system using elimination. x + 4y = 7 4x – 3y = 9 (3, 1) (3) + 4(1) = 7 4(3) - 3(1) = 9 Step 5: Check your solution.

  24. What is the first step when solving with elimination? • Add or subtract the equations. • Multiply the equations. • Plug numbers into the equation. • Solve for a variable. • Check your answer. • Determine which variable to eliminate. • Put the equations in standard form.

  25. Which variable is easier to eliminate? • x • y • 6 • 4 3x + y = 4 4x + 4y = 6

  26. 3) Solve the system using elimination. 3x + 4y = -1 4x – 3y = 7 Step 1: Put the equations in Standard Form. They already are! Find the least common multiple of each variable. LCM = 12x, LCM = 12y Which is easier to obtain? Either! I’ll pick y because the signs are already opposite. Step 2: Determine which variable to eliminate.

  27. 3) Solve the system using elimination. 3x + 4y = -1 4x – 3y = 7 Multiply both equations (3)(3x + 4y = -1) (4)(4x – 3y = 7) x = 1 9x + 12y = -3 (+) 16x – 12y = 28 Step 3: Multiply the equations and solve. 25x = 25 3(1) + 4y = -1 3 + 4y = -1 4y = -4 y = -1 Step 4: Plug back in to find the other variable.

  28. 3) Solve the system using elimination. 3x + 4y = -1 4x – 3y = 7 (1, -1) 3(1) + 4(-1) = -1 4(1) - 3(-1) = 7 Step 5: Check your solution.

  29. What is the best number to multiply the top equation by to eliminate the x’s? • -4 • -2 • 2 • 4 3x + y = 4 6x + 4y = 6

  30. Solve using elimination. • (2, 1) • (1, -2) • (5, 3) • (-1, -1) 2x – 3y = 1 x + 2y = -3

  31. Find two numbers whose sum is 18 and whose difference 22. • 14 and 4 • 20 and -2 • 24 and -6 • 30 and 8

  32. Using Elimination to Solve a Word Problem: Two angles are supplementary. The measure of one angle is 10 degrees more than three times the other. Find the measure of each angle.

  33. Using Elimination to Solve a Word Problem: Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle. x = degree measure of angle #1 y = degree measure of angle #2 Therefore x + y = 180

  34. Using Elimination to Solve a Word Problem: Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle. x + y = 180 x =10 + 3y

  35. Using Elimination to Solve a Word Problem: Solve x + y = 180 x =10 + 3y x + 42.5 = 180 x = 180 - 42.5 x = 137.5 (137.5, 42.5) x + y = 180 -(x - 3y = 10) 4y =170 y = 42.5

  36. Using Elimination to Solve a Word Problem: The sum of two numbers is 70 and their difference is 24. Find the two numbers.

  37. Using Elimination to Solve a Word problem: The sum of two numbers is 70 and their difference is 24. Find the two numbers. x = first number y = second number Therefore, x + y = 70

  38. Using Elimination to Solve a Word Problem: The sum of two numbers is 70 and their difference is 24. Find the two numbers. x + y = 70 x – y = 24

  39. Using Elimination to Solve a Word Problem: x + y =70 x - y = 24 47 + y = 70 y = 70 – 47 y = 23 2x = 94 x = 47 (47, 23)

  40. Now you Try to Solve These Problems Using Elimination. Solve • Find two numbers whose sum is 18 and whose difference is 22. • The sum of two numbers is 128 and their difference is 114. Find the numbers.

  41. Matrix Operations

  42. MATRIX: A rectangular arrangement of numbers in rows and columns. The ORDER of a matrix is the number of the rows and columns. The ENTRIES are the numbers in the matrix. What is a Matrix? This order of this matrix is a 2 x 3. columns rows

  43. What is the order? (or square matrix) 3 x 3 (Also called a column matrix) 1 x 4 3 x 5 (or square matrix) 2 x 2 4 x 1 (Also called a row matrix)

  44. Adding Two Matrices • To add two matrices, they must have the same order. To add, you simply add corresponding entries.

  45. = 7 7 4 5 = 0 7 5 7

  46. Subtracting Two Matrices To subtract two matrices, they must have the same order. You simply subtract corresponding entries.

  47. 2-0 -4-1 3-8 -5 2 -5 8-3 0-(-1) -7-1 5 -8 1 = = 1-(-4) 5-2 0-7 5 3 -7

  48. Multiplying a Matrix by a Scalar • In matrix algebra, a real number is often called a SCALAR. To multiply a matrix by a scalar, you multiply each entry in the matrix by that scalar.

  49. -3 3 -2 6 -5 -2(-3) -2(3) 6 -6 -12 -2(6) -2(-5) 10

  50. Multiplying Matrices

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