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