Solve the given system of equations.2x+3y+z=2-x+2y+3z=-1-3x-3y+z=0

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The following system of linear equations has to be solved:

2x+3y+z=2 ...(1)

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

-3x-3y+z=0 ...(3)

Add (1) and (3)

=> 2x+3y+z -3x-3y+z=2

=> -x+2z =2

=> z = (2 + x)/2

(1) - (3)

=> 2x+3y+z +3x+3y-z=2

=> 5x + 6y = 2

=> y = (2 - 5x)/6

Substitute z = (2 + x)/2 and y = (2 - 5x)/6 in (2)

-x+2*(2 - 5x)/6 + 3*(2 + x)/2 = -1

=> -x+(2 - 5x)/3 + 3*(2 + x)/2 = -1

=> -6x+6(2 - 5x)/3 + 3*6(2 + x)/2 = -6

=> -6x+2(2 - 5x) + 9(2 + x) = -6

=> -6x + 4 - 10x + 18 + 9x = -6

=> -7x = -28

=> x = 4

y = (2 - 5x)/6

=> y = -18/6 = -3

z = (2 + x)/2

=> z = 3

The solution to the set of equations is x = 4, y = -3 and z = 3

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