# 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 Monday September 29, 2014 Warm-up: Graph and shade the solution area -x + y > 4 x + y < 3

10 Monday September 29, 2014 Warm-up: Graph and shade the solution area -x + y > 4 x + y < 3

11 Monday September 29, Graphing Systems of Linear Inequalities Objective: To graph systems of linear inequalities in slopeintercept form and standard form EQ: Is graphing a system of linear inequalities any different than just a linear inequality? Steps to graph a system of inequalities 1. Write each inequalities in slope-intercept form (y=mx+b) by solving for y 2. Use b to plot y-intercept then use m to plot the 2 nd point for the 1 st inequality 3. Use dashed line for > or <. Use solid line for or 4. Shade the appropriate region 5. On the same graph, repeat steps 2,3, and 4 to plot and shade for the 2 nd inequality 6. Solution is where the graph is double shaded. 7. Check the solution algebraically Example1: Graph y > -2x -5 y x + 3 Solution: Step 1: Step 2: for line 1, m = -2 and b = -5 for line 2, m = 1 and b = 3 Step 3: line 1 is dashed, line 2 is solid Step 4: Shade the regions using check points

12 Monday September 29, 2014 Example1: Graph the system of inequalities 2x + 3y < 6 y -(2/3)x + 4 Solution: 1. Write in y=mx form (1) 3y = -2x +6 y = -(2/3) + 2 (2) y = - (2/3) x In (1), m = -(2/3), b =2 In (2), m =-(2/3), b =4 No intersection No solution 3. No solution

13 Tuesday September 30, 2014 Warm-up: Solve by elimination: x + 2y = -1 (1) 3x y = 18 (2)

14 Tuesday September 30, 2014 Warm-up: Solve by elimination: x + 2y = -1 (1) 3x y = 18 (2) Solution: 1. Multiply (1) with 3: 3x + 6y = -3-3x - y = y = -21 y = Substitute y = -3 to (2) 3x (-3) = 18 3x + 3 = 18 3x = 15 x = 5

15 3.4Solving Linear Systems in Three Variables: Objective: To solve linear systems with 3 variables EQ: How many different ways can 3 linear equations in 3 variables intersect? Steps to solve a linear system using elimination method: 1. Pick any 2 equations and eliminate 1 variable equation (4) 2. Pick 2 different equations and eliminate the same variable equation (5) 3. Use equations (4) and (5) to solve for 1 of its variables. 4. Substitute the known variable to either (4) or (5) to find the other variable. 5. Substitute the 2 known variables into any of the original equations to solve for the 3 rd variable. Example: Solve the system: x + y + z = 3 (1) 4x + 4y + 4z = 7 (2) 3x - y + 2z = 5 (3) Solution: Step 1. Pick any 2 equations and eliminate 1 variable: Multilply (1) with 4 and subtract (2) 4x + 4y + 4z = 12 (4) - 4x + 4y + 4z = 7 (2) = 5 The system has no solution Tuesday September 30, 2014

16 3.4Solving Linear Systems in Three Variables: Example: Solve the system using the elimination method 4x + 2y + 3z = 1 (1) 2x - 3y + 5z = -14 (2) 6x - y + 4z = -1 (3) Solution: Step 1. Multiply (3) with 2 and add to (1) to eliminate y 4x + 2y + 3z = 1 (1) 12x - 2y + 8z = -2 (2) x + 11z = -1 (4) Step 2. Multiply (3) with 3 and subtract to (2) to also eliminate y 2x - 3y + 5z = -14 (2) - { 18x 3y + 12z = -3 } (3) x - 7z = -11 (5) Step 3. Add (4) and (5) to eliminate x 16x + 11z = -1 (4) -16x - 7z = -11 (5) z = -12 z = -3 Use (5) to solve for x: - 16x + 7(-3) = x - 21 = x = -32 x = 2 Step4: Substitute z = -3 and x = 2 to (3) 6(2) - y + 4(-3) = -1 - y = y = -1 The solution is (2, 1, -3) Tuesday September 30, 2014

17 Example: Solve the system: x + y + z = 4 (1) x + y - z = 4 (2) 3x + 3y + z = 12 (3) Solution: Tuesday September 30, Solving Linear Systems in Three Variables: Step 1. Pick any 2 equations and eliminate 1 variable: Add (1) and (2) x + y + z = 4 (1) +{ x + y - z = 4 } (2) x +2y = 8 (4) Step2: Pick 2 different equations and also eliminate z: x + y - z = 4 (2) 3x + 3y + z = 12 (3) x + 4y = 16 (5) Step3: Use the new equations to eliminate the next variable Multiply (4) with 2 and subtract with (5) 4x + 4y = 16 4x + 4y = 16 (5) = 0 The system has infinitely many solutions.

18 3.4Solving Linear Systems in Three Variables: Example: Solve the system using the substitution method 2a + b + c = 8 (1) - a + 3b 2c = 3 (2) - a + b - c = 0 (3) Solution: Step 1. Rewrite 1 of the equations for 1 variable. - a + b - c = 0 (3) b = a + c (4) Step 2. Substitute (4) into (1) 2a + b + c = 8 (1) 2a + (a+c) +c = 8 2a + a + c + c = 8 3a + 2c = 8 (5) Step 3. Substitute (4) into (2) - a + 3b 2c = 3 (2) - a + 3(a+c) 2c = 3 - a + 3a + 3c 2c = 3 2a + c = 3 (6) Step4: Use (5) and (6) to solve for 1 variable 3a + 2c = 8 (5) 2a + c = 3 (6) Multiply (6) with 2 3a + 2c = 8 (5) 4a + 2c = 6 (6) a = 2 a = - 2 Step 5: Substitute a = -2 into (5) or (6) 2(-2) + c = c = 3 c = 7 Step 6: Substitute a = -2 and c = 7 into (3) b = a + c = = 5 The solution is (-2, 5, 7) Tuesday September 30, 2014

19 Tuesday September 30, 2014 Warm-up: Solve by elimination: x + 2y = -1 (1) 3x y = 18 (2)

20 Friday October 9, 2015 Warm-up: The school that Lisa goes to is selling tickets to the annual talent show. - On the first day of ticket sales the school sold 4 senior citizen tickets and 5 student tickets for a total of \$ The school took in \$126 on the second day by selling 7 senior citizen tickets and 5 student tickets. What is the price each of one senior citizen ticket and one student ticket?

21 Friday October 9, 2015 Warm-up: The school that Lisa goes to is selling tickets to the annual talent show. - On the first day of ticket sales the school sold 4 senior citizen tickets and 5 student tickets for a total of \$ The school took in \$126 on the second day by selling 7 senior citizen tickets and 5 student tickets. What is the price each of one senior citizen ticket and one student ticket? senior citizen ticket: \$8, student ticket: \$14

22 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 9, 2015 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.

23 Friday October 9, Matrices-Basic Operations: Scalar Multiplication Simply multiply each element with the scalar Example:

24 Warm-up: Find x and y Tuesday October 13, 2015

25 Tuesday October 13, 2015 Warm-up: Find x and y Solution: Therefore: The solution is x = -2 and y = 4

26 3.6 Multiply Matrices: Tuesday October 13, 2015 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

27 3.6 Multiply Matrices: Friday October 9, 2015 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:

28 3.6 Multiply Matrices: Monday October 19, 2015 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

29 Warm-up: Find: A(B+C) where Tuesday October 20, 2015

30 Tuesday October 20, 2015 Warm-up: Use calculator to find: A(B+C) where Solution:

31 Tuesday October 20, 2015 Warm-up: Multiply. Give reason why if not possible.

32 Tuesday October 20, 2015 Warm-up: Multiply. Give reason why if it s not.

33 Wednesday October 21, Inverse Matrices: Objective: To use inverse matrices to solve linear systems EQ: How can we solve linear systems using matrices? Definition: Identity matrix: a matrix with 1 s on the main diagonal and 0 s Inverse Matrices: Matrices A and B are inverses of each other if their product, AB and BA, is an identity matrix. To find the inverse of matrix A Example: Solution: The inverse of a 3x3 matrix is difficult to compute by hand. Instead, for matrices that greater then 2x2, use calculator.

34 3.8 Inverse Matrices: Wednesday October 21, 2015 Using an Inverse Matrix to Solve a Linear System: Example: Solution:

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