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Do Now 1/12/10 Take Text p. 430, #4-20, 30-34 evens Text p. 439, #4-24 even, #32, & #36 Copy out HW from last night. HW in your planner. Text p. 441, #1-9 all Text p. 447, #6-28 evens Homework Text p. 430, #4-20, 30-34 evens 4) not a solution 6) B 8) (1,-3) 10) (3,2) 12) (1,2) 14) (-2,-2) 16) (1.8, -0.4) 18) (4,3) 20) (1, -2) 30) (3,5), (1,2), & (5,1) 32) B 34a) 25 bike, 15 stairs b) 25 elliptical trainer and 5 stairs Homework Text p. 439, #4-24, 32 & 36 evens 4) (1,1) 6) (13,6) 8) (3,22) 10) (8,-7) 12) (-1,11) 14) (4,3) 16) (-1,1) 18) A 20) (5,1) 22) (10.5, 11.75) 24) (14, 10) 32) 22 tubes, 4 “cooler tubes” 36) 12 quarters Section 7.1 “Solve Linear Systems by Graphing” Linear System– consists of two more linear equations. x + 2y = 7 3x – 2y = 5 Equation 1 Equation 2 A solution to a linear system is an ordered pair (a point) where the two linear equations (lines) intersect (cross). Section 7.2 “Solve Linear Systems by Substitution” Solving a Linear System by Substitution (1) Solve one of the equations for one of its variables. (When possible, solve for a variable that has a coefficient of 1 or -1). (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. “Solve Linear Systems by Substituting” Equation 1 x – 2y = -6 Equation 2 4x + 6y = 4 x = -6 + 2y 4(-6+ +6y2y) 4x = 4+ 6y = 4 -24 + 8y + 6y = 4 -24 + 14y = 4 y =2 x – 2y = -6 Equation 1 x = -6 + 2(2) x = -2 (-2) - 2(2) = -6 -6 = -6 Substitute Substitute value for x into the original equation The solution is the point (-2,2). Substitute (-2,2) into both equations to check. 4(-2) + 6(2) = 4 4=4 “How Do You Solve a Linear System???” (1) Solve Linear Systems by Graphing (7.1) (2) Solve Linear Systems by Substitution (7.2) (3) Solve Linear Systems by ELIMINATION!!! (7.3) Adding or Subtracting Section 7.3 “Solve Linear Systems by Adding or Subtracting” ELIMINATIONadding or subtracting equations to obtain a new equation in one variable. Solving Linear Systems Using Elimination (1) Add or Subtract the equations to eliminate one variable. (2) Solve the resulting equation for the other variable. (3) Substitute in either original equation to find the value of the eliminated variable. “Solve Linear Systems by Elimination” Equation 1 Equation 2 + 2x + 3y = 11 -2x + 5y = 13 8y = 24 y=3 2x + 3y = 11 Equation 1 2x + 3(3) = 11 2x + 9 = 11 x=1 2(1) + 3(3) = 11 11 = 11 Substitute value for y into either of the original equations The solution is the point (1,3). Substitute (1,3) into both equations to check. -2(1) + 5(3) = 13 13 = 13 “Solve Linear Systems by Elimination” Equation 1 Equation 2 _ + 4x + 3y = 2 -5x 5x + -3y 3y ==-2 2 -x 4x + 3y = 2 Equation 1 4(-4) + 3y = 2 -16 + 3y = 2 y=6 4(-4) + 3(6) = 2 2=2 = 4 x = -4 Substitute value for x into either of the original equations The solution is the point (-4,6). Substitute (-4,6) into both equations to check. 5(-4) + 3(6) = -2 -2 = -2 “Solve Linear Systems by Elimination” Equation 1 Equation 2 + 8x - 4y = -4 -3x 4y + = 4y 3x = + 14 5x 8x - 4y = -4 Equation 1 8(2) - 4y = -4 16 - 4y = -4 y=5 8(2) - 4(5) = -2 -2 = -2 Arrange like terms = 10 x= 2 Substitute value for x into either of the original equations The solution is the point (2,5). Substitute (2,5) into both equations to check. 4(5) = 3(2) + 14 20 = 20 Guided Practice 4x – 3y = 5 -2x + 3y = -7 7x – 2y = 5 7x – 3y = 4 (-1, -3) (1,1) 3x + 4y = -6 2y = 3x + 6 (-2,0) NJASK7 prep Homework Text p. 447, #6-28 evens