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Linear Systems in Three or More Variables

Linear Systems in Three or More Variables. (teacherweb.com). Solve using back-substitution. x – 2y + 3z = 9 y + 3z = 5 z = 2. Sub. y = -1 and z = 2 into 1 st equation. Sub. z = 2 into 2 nd equation. y + 3(2) = 5 y + 6 = 5 y = -1. x – 2(-1) + 3(2) = 9

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Linear Systems in Three or More Variables

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  1. Linear Systems in Three or More Variables (teacherweb.com)

  2. Solve using back-substitution. x – 2y + 3z = 9 y + 3z = 5 z = 2 Sub. y = -1 and z = 2 into 1st equation. Sub. z = 2 into 2nd equation. y + 3(2) = 5 y + 6 = 5 y = -1 x – 2(-1) + 3(2) = 9 x + 2 + 6 = 9 x + 8 = 9 x = 1 Answer (x, y, z ) = (1, -1, 2)

  3. Objective - To solve systems of linear equations in three variables. Solve.

  4. Describe all the ways that three planes could intersect in space. Intersects at a Point One Solution

  5. Describe all the ways that three planes could intersect in space. Intersects at a Line Infinitely Many Solutions

  6. Describe all the ways that three planes could intersect in space. No Solution

  7. Describe all the ways that three planes could intersect in space. No Solution

  8. Solve.

  9. Solve.

  10. Solve. IdentityInfinitely Many Solutions

  11. In 1998, Cynthia Cooper of the WNBA Houston Comets basketball team was named Team Sportswoman of the Year. Cooper scored 680 points by hitting 413 of her 1-pt., 2-pt. and 3-point attempts. She made 40% of her 160 3-pt. field goal attempts. How many 1-, 2- and 3-point baskets did Ms. Cooper make? x = number of 1-pt. free throws y = number of 2-pt. field goals z = number of 3-pt. field goals x + y + z = 413 x + 2y + 3z = 680 z/160 = 0.4 -x - y - z = -413 x + 2y + 3z = 680 y + 2z = 267 y + 2(64) = 267 y = 139 x + 139 + 64 = 413 x = 210 z = 64

  12. Find a quadratic function f(x) = ax2 + bx + c the graph of which passes through the points (-1, 3), (1, 1), and (2, 6). Plug in each point for x and y. a(-1)2 + b(-1) + c = 3 a(1)2 + b(1) + c = 1 a(2)2 + b(2) + c = 6 Simplify a – b + c = 3 a + b + c = 1 4a + 2b + c = 6

  13. Find a quadratic function f(x) = ax2 + bx + c the graph of which passes through the points (-1, 3), (1, 1), and (2, 6). a – b + c = 3 a + b + c = 1 -2a - 2b - 2c = -2 4a + 2b + c = 6 a – b + c = 3 a + b + c = 1 4a + 2b + c = 6 2a + 2c = 4 2a – c = 4 2a – c = 4 2a + 2c = 4 -2a + c = -4 2a + 2c = 4 a – b + 0 = 3 a + b + 0 = 1 a – b = 3 a + b = 1 3c = 0 c = 0 2 + b + 0 = 1 b = -1 2a = 4 a = 2 f(x) = 2x2 – x

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