Put the following equations to slope-intercept form then use 2 points to graph

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1 Tuesday September 23, 2014 Warm-up: Put the following equations to slope-intercept form then use 2 points to graph 1. 4x - 3y = 8 8 x 6y = x + y = 4 2x + y = 1

2 Tuesday September 23, 2014 Warm-up: Put the following equations to slope-intercept form then use 2 points to graph 1. 4x - 3y = 8 8 x 6y = x + y = 4 2x + y = 1

3 Tuesday September 23, Solving Linear Systems using Graph Objective: To solve a system of Linear equations EQ: How many ways can we solve a system of linear equations Definition: A system of linear equations consists of 2 or more equations The solution of a system of linear equations can be 1 of the following 3 cases: 1. Exactly 1 solution Example1: When slope (m) and y-intercept (b) are given Lines intersect at 1 point Write the equation for the line in (consistent the graph and independent) Solution: Use Slope-Intercept form: According 2. Infinitely to the graph, many b solutions = -1 and m = ( ) / (3 0) = -2/3 Lines coincide Therefore, the equation is y = (-2/3) x - 1 (consistent and dependent) 3. No solutions Lines are parallel (inconsistent)

4 3.1 Solving Linear Systems using Graph There are 4 ways to solve a system of linear equations 1. Graphing 2. Substitution 3. Elimination 4. Multiplication/Elimination Steps to solve a linear system using graphing: 1. Put each equation in slope-intercept form (y = mx+b) 2. Use y-intercept to plot 1 st point and use m to plot 2 nd point 3. Look for solution(s) 4. Substitute the solution in each equation to check for error Example: Graph the linear system and estimate the solution then check the solution algebraically 4x + y = 8 (1) 2x 3y = 18 (2) Solution: 1. Put equations in y = mx + b form y = -4x + 8 (1) y = (2/3)x - 6 (2) 2. In (1), m = -4, b = 8 In (2), m = 2/3, b = 6 3. Solution is (3, -4) 4. Check: Tuesday September 23, 2014

5 Warm-up: Wednesday September 24, 2014

6 Example: Solve the system using the substitution method 2x + 5y = -5 (1) x + 3y = 3 (2) Solution: Step 1. Solve (2) for x x + 3y = 3 (2) x = -3y + 3 (3) Step 2. Substitute (3) into (1) 2x + 5y = -5 (1) 2 (-3y +3) + 5 y = -5-6 y y = y = -5 y = 11 Wednesday September 24, Solving Linear Systems by Substitution Steps to solve a linear system using substitution method: 1. Solve one of the equation for one of its variables 2. Substitute the expression from step 1 into the other equation and solve for the other variable 3. Substitute the value from step 2 into the revised equation from step 1 and solve for the other variable Step 3. Substitute y =11 back to either (1) or (2) x + 3y = 3 (2) x + 3 (11) = 3 x + 33 = 3 x = -30 The solution is (-30, 11) Check:

7 Example: Solve the system using the elimination method 6x - 14y = 20 (1) 6x - 8y = 8 (2) Solution: Step 1. Subtract (1) and (2) to eliminate x 6x - 14y = 20 (1) 6x - 8y = 8 (2) y = 12 y = - 2 Wednesday September 24, Solving Linear Systems by Elimination Steps to solve a linear system using elimination method: 1. Add or subtract the equations to eliminate one of the variables. Then solve for the other variable 2. Substitute the expression from step 1 into either the original equation and solve for the other variable. Step 2. Substitute y = -2 into either (1) or (2) 6x - 8y = 8 (2) 6x - 8 (-2) = 8 6x + 16 = 8 6x = x = -8 x = -8/6 = -4/3 The solution is (-4/3, -2) Check: (1) 6(-4/3) - 14 (-2) = 20 (2) 6(-4/3) 8 (-2) = = = 8 Yes Yes

8 Wednesday September 24, Solving Linear Systems by Multiplication and Elimination Steps to solve a linear system using multiplication and elimination method: 1. Multiply on or both of the equation by a constant to make the coefficient of one variable in both equations the same 2. Add or subtract the equations to eliminate one of the variables. Then solve for the other variable 3. Substitute the expression from step 1 into either the original equation and solve for the other variable. Example: Solve the system using the elimination method 3x - 7y = 10 (1) 6x - 8y = 8 (2) Solution: Step 1. Multiply (1) with 2 2 (3x - 7y = 10) x - 14y = 20 Step 2. Subtract (1) and (2) to eliminate x 6x - 14y = 20 (1) 6x - 8y = 8 (2) y = 12 y = - 2 Step 2. Substitute y = -2 into either (1) or (2) 6x - 8y = 8 (2) 6x - 8 (-2) = 8 6x + 16 = 8 6x = x = -8 x = -8/6 = -4/3 The solution is (-4/3, -2) Check: (1) 6(-4/3) - 14 (-2) = 20 (2) 6(-4/3) 8 (-2) = = = 8 Yes Yes

9 3.5 Matrices-Basic Operations: Matrix is a rectangular arrangement of numbers in row and columns. Dimension: m x n (row x column) Each number is an element Two matrices are equal iff: 1. Their dimensions are the same 2. Corresponding elements are equal Friday October 18, 2013 Objective: To perform basic operations with matrices EQ: How can we organize sports data? Adding and Subtracting Matrices (only when dimensions are the same) Simply add or subtract corresponding elements in the corresponding positions Example: 1. 2.

10 Friday October 18, Matrices-Basic Operations: Scalar Multiplication Simply multiply each element with the scalar Example:

11 Warm-up: Find x and y Monday October 20, 2014

12 Monday October 20, 2014 Warm-up: Find x and y Solution: Therefore: The solution is x = -2 and y = 4

13 3.6 Multiply Matrices: Monday October 20, 2013 Objective: To perform basic operations with matrices EQ: How can we organize sports data? The product of 2 matrices A and B is defined iff: the number of columns in A is equal to the number of rows in B Example: State whether the product AB is defined: 1. A: 5x2 and B:2x2 defined, AB: 5x2 3. A: 4x3 and B:3x2 defined, AB: 4x2 2. A: 3x2 and B:3x2 undefined 4. A: 3x4 and B:3x2 undefined

14 3.6 Multiply Matrices: Monday October 21, 2013 Multiplying Matrices Multiply each element in the ith row of A to jth column of B Note: The product AB BA Example: Find product AB then find product BA if Solution: Find AB: since A: 2x2 and B: 2x2, AB: 2x2 1. AB = 2. AB = 3. AB = 4. AB = 5. AB = Find BA:

15 3.6 Multiply Matrices: Monday October 20, 2013 Example: Use matrices to calculate the total cost of 2 hockey teams. Sticks ($60) Pucks ($2) Women steam Men s Team Uniforms ($35) Solution: Let s write the equipment list and the cost list in matrix form: So, the total cost of equipment for each team can be found by multiply the equipment matrix (E) by the cost matrix (C ). Since E: 2x3 and C: 3x1, then EC: 2x1 The total cost for women s team is $1530 The total cost for men s team is $1710

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