 # Do Now 1/12/12  In your notebook, answer the following question. During a football game, a bag of popcorn sells for \$2.50 and a pretzel sells for \$2.00.

## Presentation on theme: "Do Now 1/12/12  In your notebook, answer the following question. During a football game, a bag of popcorn sells for \$2.50 and a pretzel sells for \$2.00."— Presentation transcript:

Do Now 1/12/12  In your notebook, answer the following question. During a football game, a bag of popcorn sells for \$2.50 and a pretzel sells for \$2.00. The total amount of money collected during the game was \$336. Twice as many bags of popcorn sold compared to pretzels. How many bags of popcorn and pretzels were sold during the game? x = y = Mid-Term Review

During a football game, a bag of popcorn sells for \$2.50 and a pretzel sells for \$2.00. The total amount of money collected during the game was \$336. Twice as many bags of popcorn sold compared to pretzels. How many bags of popcorn and pretzels were sold during the game? During a football game, a bag of popcorn sells for \$2.50 and a pretzel sells for \$2.00. The total amount of money collected during the game was \$336. Twice as many bags of popcorn sold compared to pretzels. How many bags of popcorn and pretzels were sold during the game? 96 bags of popcorn and 48 pretzels x = y = y = 2x \$2.50y + \$2.00x = \$336

Linear System– consists of two more linear equations. consists of two more linear equations. Equation 1 Equation 1 3x – 2y = 5 3x – 2y = 5 Equation 2 Equation 2 x + 2y = 7 x + 2y = 7 Section 7.1 “Solve Linear Systems by Graphing” A solution to a linear system is an ordered pair (a point) where the two linear equations (lines) intersect (cross).

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. Section 7.2 “Solve Linear Systems by Substitution”

Equation 1 Equation 1 4x + 6y = 4 4x + 6y = 4 Equation 2 Equation 2 x – 2y = -6 x – 2y = -6 “Solve Linear Systems by Substituting” 4x + 6y = 4 4x + 6y = 4 4(-6 + 2y) + 6y = 4 4(-6 + 2y) + 6y = 4 Substitute -24 + 8y + 6y = 4 -24 + 8y + 6y = 4 -24 + 14y = 4 -24 + 14y = 4 y = 2 y = 2 Equation 1 Equation 1 x – 2y = -6 x – 2y = -6 Substitute value for x into the original equation x = -6 + 2(2) x = -6 + 2(2) x = -2 x = -2 The solution is the point (-2,2). Substitute (-2,2) into both equations to check. 4(-2) + 6(2) = 4 4(-2) + 6(2) = 4 4 = 4 (-2) - 2(2) = -6 (-2) - 2(2) = -6 -6 = -6 x = -6 + 2y x = -6 + 2y

“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”  ELIMINATION- adding 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.

Equation 1 Equation 1 -2x + 5y = 13 -2x + 5y = 13 Equation 2 Equation 2 2x + 3y = 11 2x + 3y = 11 “Solve Linear Systems by Elimination” Equation 1 Equation 1 2x + 3y = 11 2x + 3y = 11 Substitute value for y into either of the original equations 2x + 3(3) = 11 2x + 3(3) = 11 2x + 9 = 11 2x + 9 = 11 The solution is the point (1,3). Substitute (1,3) into both equations to check. -2(1) + 5(3) = 13 -2(1) + 5(3) = 13 13 = 13 2(1) + 3(3) = 11 2(1) + 3(3) = 11 11 = 11 ADDITION + 8y = 24 8y = 24 y = 3 y = 3 x = 1 x = 1 Eliminated

Equation 1 Equation 1 5x + 3y = -2 5x + 3y = -2 Equation 2 Equation 2 4x + 3y = 2 4x + 3y = 2 “Solve Linear Systems by Elimination” Equation 1 Equation 1 4x + 3y = 2 4x + 3y = 2 Substitute value for x into either of the original equations 4(-4) + 3y = 2 4(-4) + 3y = 2 -16 + 3y = 2 -16 + 3y = 2 The solution is the point (-4,6). Substitute (-4,6) into both equations to check. 5(-4) + 3(6) = -2 5(-4) + 3(6) = -2 -2 = -2 4(-4) + 3(6) = 2 4(-4) + 3(6) = 2 2 = 2 SUBTRACTION _ -x = 4 -x = 4 x = -4 x = -4 y = 6 y = 6 Eliminated -5x + -3y = 2 -5x + -3y = 2 +

Equation 1 Equation 1 4y = 3x + 14 4y = 3x + 14 Equation 2 Equation 2 8x - 4y = -4 8x - 4y = -4 “Solve Linear Systems by Elimination” Equation 1 Equation 1 8x - 4y = -4 8x - 4y = -4 Substitute value for x into either of the original equations 8(2) - 4y = -4 8(2) - 4y = -4 16 - 4y = -4 16 - 4y = -4 The solution is the point (2,5). Substitute (2,5) into both equations to check. 4(5) = 3(2) + 14 4(5) = 3(2) + 14 20 = 20 8(2) - 4(5) = -2 8(2) - 4(5) = -2 -2 = -2 + 5x = 10 5x = 10 x = 2 x = 2 y = 5 y = 5 Eliminated -3x + 4y = 14 -3x + 4y = 14 Arrange like terms

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)

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