 Describe each transformation of f to g: f(x) = |x| and g(x) = -3|x + 2| - 1.  Write a function g if f(x) = x 2 has a vertical shrink of 1/3 followed by a translation up 2.  The data shows the humerus lengths ( in centimeters) and heights (in centimeters( of several females. Use the graphing calculator to find a line of best fit for the data. Estimate the height of a female whose humerus is 40 centimeters long. Estimate the humerus length of a female with a height of 130 cm. Algebra II 1

Systems of Equations with Two Variables Algebra II

 two or more linear equations.  Looks like  A solution is an ordered pair that makes all equations true. Algebra II 3

3x – 2y = 2 x + 2y = 6 a) (0, -1) b) (2,2) no yes Algebra II 4

 Graphing  Substitution  Elimination 5 Algebra II

To find the solution of a system of two linear equations: (steps) 1. Graph each equation 2. Identify the intersection 3. This is the solution to the system because it is the point that satisfies both equations. **Remember that a graph is just a picture of the solutions. Algebra II 6

coincident lines (same line) infinitely many solutions parallel lines no solutions intersecting lines one solution Graph Number of Solutions Two lines intersect at one point. Parallel lines Lines coincide Algebra II 7

First, graph 2x – 2y = -8. Second, graph 2x + 2y = 4. The lines intersect at (-1, 3) The solution is (-1, 3) Solve the system of equations by graphing. Algebra II 8

First, graph -x + 3y = 6. (0, -1) (3, 0) Second, graph 3x – 9y = 9. (0, 2) (-3, 1) The lines are parallel. No solution (3, 3) Solve the system of equations by graphing. (-3, -2) Algebra II 9

First, graph 2x – y = 6. Second, graph x + 3y = 10. The lines intersect at (4, 2) The solution is (4, 2) Solve the system of equations by graphing. Algebra II 10

First, graph x = 3y – 1. (-1, 0) (2, 1) Second, graph 2x – 6y = -2. The lines are identical. Infinitely many solutions Solve the system of equations by graphing. (-4, -1) Algebra II 11

Steps for Substitution: 1. Solve one of the equations for one variable (try to solve for the variable with a coefficient of one) 2. Substitute the expression into the other equation and solve the new equation. 3. Substitute the value from step 2 into one of your original equations to complete the ordered pair Algebra II 12

1. 3x – y = 6 -4x + 2y = –8 Step 1: 3x – y = 6 -y = -3x + 6 y = 3x – 6 Step 2: -4x + 2y = -8 -4x + 2(3x – 6) = -8 -4x + 6x – 12 = -8 2x = 4 x = 2 Step 3: y = 3x – 6 y = 3(2) – 6 y = 0 Algebra II 13 (2,0)

2. x – 3y = 4 6x – 2y = 4 Step 1: x – 3y = 4 x = 3y +4 Step 2: 6x – 2y = 4 6(3y + 4) – 2y = 4 18y + 24 – 2y = 4 16y = -20 y = -5/4 Step 3: x = 3y + 4 x = 3(-5/4) + 4 x = 1/4 Algebra II 14 (1/4, -5/4)

3. y = 2x – 5 8x – 4y = 20 Step 1: Y = 2x – 5 (already done) Step 2: 8x – 4y = 20 8x – 4(2x – 5) = 20 8x – 8x + 20 = = 20 0 = 0 True Statement! Algebra II 15 Infinitely Many Solutions

4. -4x + y = 6 -5x – y = 21 Step 1: -4x + y = 6 y = 4x + 6 Step 2: -5x – y = 21 -5x – (4x + 6) = 21 -5x – 4x – 6 = 21 -9x = 27 x = -3 Step 3: y = 4x + 6 y = 4(-3) + 6 y = -6 Algebra II 16 (-3,-6)

 Steps for elimination: 1. Make one of the variables have opposite coefficients (multiply by a constant if necessary) 2. Add the equations together and solve for the remaining variable 3. Substitute the value from step 3 into one of the original equations to complete the ordered pair Algebra II 17

Solve the following system by elimination 6x – 3y = –3 4x + 5y = –9 30x – 15y = x + 15y = x + 0 = x = x = (6x – 3y = –3) 3(4x + 5y = –9) Algebra II 18

Use x = -1 to find y 2 nd equation : 4x + 5y = -9 4(-1) + 5y = y = y = -5 5 y = -1 (-1, -1) 3 Algebra II 19

Solve the following system by elimination 3x – y = 4 6x – 2y = 4 -6x + 2y = -8 6x – 2y = = -4 0 = -4 False! No Solution 12 -2(3x – y = 4) (6x – 2y = 4) ≠ Algebra II 20

Solve the following system by elimination 3x + 5y = -6 2x – 2y = -8 6x + 10y = x + 6y = y = 12 16y = y = 3/4 12 2(3x + 5y = -6) -3(2x – 2y = -8) Algebra II 21

Use y = 3/4 to find x 1 st equation : 3x + 5y = -6 3x+ 5(3/4) = -6 3x + 15/4 = /4 -15/4 3x = -39/4 3 3 y = -13/4 (-13/4, 3/4) 3 Algebra II 22

Solve the following system by elimination -2x + y = -5 8x – 4y = 20 -8x + 4y = -20 8x – 4y = = 0 0 = 0 True! Infinitely Many Solutions 12 4(-2x + y = -5) (8x – 4y = 20) = Algebra II 23

24 Algebra II 1. 4x – 3y = 10 2x + 2y = 7 2. Y = 3x – 5 2x + 3y = 8 3. X – 3y = 10 4x + 3y = x + 2y = 8 2y + 4x = x + 7y = 10 x + 4y = 9 6. x – 3y = -6 x = 2y

1. 4x – 3y = 10 8x – 6y = x + 3y = 10 2x – 2y = 15 M = 4/3, b= -10/3 M = 4/3 b = -5/6 No solution M = -1, b = 10/3 M = 1, b = -15/2 One solution Algebra II 25

3. y = 2x + 8 2x – y = /2x + 3y = 6 1/3x – 5y = -3 M = 2, b= 8 Infinitely many M = -1/6, b = 2 M = 1/15, b = 3/5 One solution Algebra II 26

1. Your family is planning a 7 day trip to Florida. You estimate that it will cost $275 per day in Tampa and $400 per day in Orlando. Your total budget for the 7 day is $2300. How many days should you spend in each location?  X = # of days in Tampa  Y = # of days in Orlando  X + y = 7  275x y = Algebra II

2. You plan to work 200 hours this summer mowing lawns or babysitting. You need to make a total of $1300. Babysitting pays $6 per hour and lawn mowing pays $8 per hour. How many hours should you work at each job?  X = # of hours babysitting  Y = # of hours of mowing  X + y = 200  6x + 8y = Algebra II

3. You make small wreaths and large wreaths to sell at a craft fair. Small wreaths sell for $8 and large wreaths sell for $12. You think you can sell 40 wreaths all together and want to make $400. How many of each type of wreath should you bring to the fair?  X = # small wreaths  Y = # large wreaths  X + y = 40  8x + 12y = Algebra II

4. You are buying lotions or soaps for 12 of your friends. You spent $100. Soaps cost $5 a piece and lotions are $8. How many of each did you buy? x = # of soaps y = # of lotions x + y = 12 5x + 8y = 100 Algebra II 30

5. Becky has 52 coins in nickels and dimes. She has a total of $4.65. How many of each coin does she have? x = # of nickels y = # of dimes x + y = 52.05x +.10y = 4.65 Algebra II 31

6. There were twice as many students as adults at the ball game. There were 2500 people at the game. How many students and how many parents were at the game? x = # of students y = # of parents x = 2y x + y = 2500 Algebra II 32

1. Using substitution, solve the system: 3x + 4y = -4 x + 2y = 2 2. Using elimination, solve the system: -3x + y = 11 5x – 2y = -16 { { (-6, -7) (-8, 5) Algebra II 33