Question 205292: solve the system of equations
3x+4y+z=7
2y+z=3
-5+3y+8z=-31
i need you help, im desperate.
Answer by alicealc(293)   (Show Source):
You can put this solution on YOUR website! I there should be an 'x' in your 3rd equation, so it will be -5x + 3y + 8z = -31
eliminate x from the 1st and 3rd equation by
multiply the 1st equation by 5:
15x + 20y + 5z = 35
and multiply the 3rd equation by 3:
-15x + 9y + 24z = -93
then add them up:
15x + 20y + 5z = 35
-15x + 9y + 24z = -93
------------------------ (+)
29y + 29z = -58
divide both sides of this new equation by 29:
y + z = -2
subtract this new equation by 2nd equation:
y + z = -2
2y + z = 3
------------ (-)
-y = -5
y = 5
substitute y = 5 into y + z = -2
5 + z = -2
z = -2 - 5 = -7
substitute y = 5, z= -7 into the 1st equation:
3x + 4y + z = 7
3x + 4*5 + (-7) = 7
3x + 20 - 7 = 7
3x + 13 = 7
3x = 7 - 13
3x = -6
x = -6/3 = -2
so,the solution is: x = -2, y = 5, z = -7
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