SOLUTION: I need to determine x, y and z when x+y+z=1150 and x=4z-100 and x=6y+50. So far the only thing I have figured out is that 1150=1150. Not a good answer. Can someone please he

Question 168385: I need to determine x, y and z when x+y+z=1150 and x=4z-100 and x=6y+50. So far the only thing I have figured out is that 1150=1150. Not a good answer.
Can someone please help.
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
determine x, y and z when
x+y+z=1150
x=4z-100
x=6y+50
-------------
x + y + z = 1150
x + 0 -4z = -100
x -6y + 0 = 50
--------------------
Use any method to solve the system to get:
x = 800
y = 125
z = 225
=================
Cheers,
Stan H.
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
There are 3 equations and 3 unknowns, so
it should be solvable
(1) given
(2) given
(3)
(4)
(5) given
(6)
(7)
Now substitute (4) and (7) into (1)
(8)
Multiply both sides by
(9)
(10)

And from (4)
(4)
(11)
(12)
And from (7)
(7)
(13)
(14)
(15)
(16)
The answers are x = 800, y = 125, z = 225
check answers:
(1) given
(2) given
(5) given
--------------------------------
(1)
(17)
(18)
--------------------------------
(2)
(19)
(20)
--------------------------------
(5) given
(21)
(22)
OK
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