1.7 - Exploring Linear Systems

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Presentation transcript:

1.7 - Exploring Linear Systems B C 2x + 3y = -4 -4x – 3y = -1 2y = 6 – 3x 6x – 5 = -4y x – y = 5 3x = 15 + 3y How many solutions can a linear system have? How can you predict the number of solutions without solving the system? A B C Linear System 2x + 3y = -4 -4x – 3y = -1 2y = 6 – 3x 6x – 5 = -4y x – y = 5 3x = 15 + 3y Number of Solutions ?

Case A One point of intersection, so 1 solution. 2x + 3y = -4 -4x – 3y = -1 One point of intersection, so 1 solution. The coefficients and constants are not multiplied by the same amount.

Case B No points of intersection, so no solutions. 2y = 6 – 3x 6x – 5 = -4y Case B No points of intersection, so no solutions. The coefficients are multiplied by the same amount, but the constants are not.

C x – y = 5 3x = 15 + 3y Case C 2 lines overlap, so every point in one line intersects with the other line. Infinite number of solutions Coefficients and constants are multiplied by the same amount.

In Summary… A linear system can have no solution, one solution, or an infinite number of solutions.