Question

ALGEBRA 2 Solve this system of equations for z, or put no solution or infinitely many.

Answer 1

Answer:

(x, y, z) = (4, -3, 2)

Step-by-step explanation:

I. -2x + y + 6z = 1

II. 3x + 2y + 5z = 16

III. 7x + 3y - 4z = 11

use I and II. cancel out a variable

-2(-2x + y + 6z = 1)

3x + 2y + 5z = 16

+ (4x - 2y - 12z = -2)

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IV. 7x - 7z = 14

use I and III. cancel out the same variable (y)

-3(-2x + y + 6z = 1)

7x + 3y - 4z = 11

+ (6x - 3y - 18z = -3)

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V. 13x - 22z = 8

use IV and V. cancel out a variable

13(7x - 7z = 14)

-7(13x - 22z = 8)

91x - 91z = 182

+(-91x + 154z = -56)

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63z = 126

divide by 63

z = 2

plug z into IV

7x - 7z = 14

7x - 7(2) = 14

7x - 14 = 14

7x = 28

x = 4

plug x and z into I

-2x + y + 6z = 1

-2(4) + y + 6(2) = 1

-8 + y + 12 = 1

y + 4 = 1

y = -3

(x, y, z) = (4, -3, 2)