Solving Systems of Equations By Substitution

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Presentation transcript:

Solving Systems of Equations By Substitution A-REI.3; A-REI.5; A-REI.6; A-REI.7

Table of Contents Pg Title LI/SC Foldable Substitution Method Notes

Learning Intention/Success Criteria LI: We are learning how to solve a system of equations by substitution SC: I know how to -determine if a system of equations has many solutions -determine if a system of equations has no solutions -determine if a system of equations has one solution -solve systems of two linear equations algebraically using substitution method -solve equations for a variable

Example 1: Solve the system of equations using substitution method 3x + y = 2 y = 2x – 3 3x + 1(2x – 3) = 2 3x + 2x – 3 = 2 5x – 3 = 2 + 3 + 3 __________________ 5x = 5 ___ 5 ___ 5 x = 1

x = 1 y = 2x – 3 y = 2(1) – 3 y = 2 – 3 y = – 1 Solution: (1, -1)

Example 2: Solve the system of equations using substitution method x – 2y = 3 x – 2y = -4 _______________ x = 2y + 3 x = 2y + 3 x – 2y = -4 1(2y + 3) – 2y = -4 2y + 3 – 2y = -4 0y + 3 = -4 3 = -4 No Solution

Example 3: Solve the system of equations using substitution method y = 2x + 1 2y = 4x + 2 2(2x + 1) = 4x + 2 4x + 2 = 4x + 2 - 2 - 2 ___________________ 4x = 4x Many solutions