Question 181780: Hi there my question is:
Solve for (x, y, z)
2x + 5y + 3z = 16
x/5 + y/2 + z/3 = 2 (this is a fraction with x at the top and 5 and the bottom)
3x = 2y - 4z = -2
State clearly whether this system is consistent or inconsistent, whether the solution is unique, or if there are an infinite number of solutions, or no solution at all. If the solution is unique, state the solution.
Thanks tutors!!
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Solve for (x, y, z)
2x + 5y + 3z = 16
x/5 + y/2 + z/3 = 2
3x - 2y - 4z = -2 (assumed that 1st equal sign is a minus)
State clearly whether this system is consistent or inconsistent, whether the solution is unique, or if there are an infinite number of solutions, or no solution at all. If the solution is unique, state the solution.
:
Multiply the 2nd equation by 30 to get rid of the denominators, we then have:
2x + 5y + 3z = 16
6x +15y +10z = 60
3x - 2y - 4z = -2
:
multiply 3rd equation by 2 and subtract from the 2nd equation
6x +15y +10z = 60
6x - 4y - 8z = -4
----------------------subtraction eliminates x:
+19y + 18z = 64
:
Multiply the 1st equation by 3, subtract from the 2nd equation:
6x +15y +10z = 60
6x +15y + 9z = 48
--------------------subtraction eliminates x & y, find z:
z = 12
;
Substitute 12 for z in the equation: +19y + 18z = 64
19y + 18(12) = 64
19y + 216 = 64
19y = 64 - 216
19y = -152
y = 
y = -8
:
Find x using the 1st equation, substitute for y & z:
2x + 5y + 3z = 16
2x + 5(-8) + 3(12) = 16
2x - 40 + 36 = 16
2x - 4 = 16
2x = 20
x = +10
:
Obviously a unique solution: x=10, y=-8, z=12
:
You can check the solutions in the 2nd original equation.
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