Solving Systems Using Elimination

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Presentation transcript:

Solving Systems Using Elimination Mrs. Book Liberty Hill Middle School Algebra I

Elimination Method You can use the Addition and Subtraction Properties of Equality to solve a system by the elimination method. Addition Property of Equality: If a = b, then a + c = b + c. Subtraction Property of Equality: If a = b, then a – c = b – c.

Adding Equations Solve by elimination: 5x – 6y = -32 3x + 6y = 48 Step 1: Eliminate y because the sum of the coefficients of y is zero.

Adding Equations 5x – 6y = -32 3x + 6y = 48 8x + 0 = 16 x = 2

Adding Equations Solve by elimination: 5x – 6y = -32 3x + 6y = 48 Step 2: Solve for the eliminated variable y using either of the original equations.

Adding Equations 3x + 6y = 48 3(2) + 6y = 48 6 + 6y = 48 y = 7 The solution is (2, 7)

Solve by elimination 6x – 3y = 3 -6x + 5y = 3

Subtracting Equations y = 0.5x + 2 1.5x + y = 42

Multiplying One Equation Solve by the elimination method: 2x + 5y = -22 10 x + 3y = 22 Step 1: Eliminate one variable To prepare for eliminating x, multiply the first equation by 5 5(2x + 5y = -22) 10x + 3y = 22

Multiplying One Equation 10x + 25y = -110 10x + 3y = 22 Subtract the equations to eliminate x. -( 10x + 3y = 22) 0x + 22y = -132 y = -6

Multiplying One Equation Solve for the eliminated variable using either of the original equations. 2x + 5y = -22 2x + 5(-6) = -22 2x – 30 = -22 2x = 8 x = 4 The solution is (4, -6)

Multiplying One Equation Solve by elimination -2x + 15y = -32 7x - 5y = 17

Multiplying Both Equations Solve by elimination 4x + 2y = 14 7x – 3y = -8 Step 1: Eliminate one variable Eliminate the variable y by multiplying the first equation by 3 and the second equation by 2.

Multiplying Both Equations 4x + 2y = 14 7x – 3y = -8

Solve by elimination 15x + 3y = 9 10x + 7y = -4