SOLUTION: solve for x, y, and z in the following systems of three equations:
10x+y+z=603
8x+2y+z=603
20x-10y-2z=-6
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Question 262256: solve for x, y, and z in the following systems of three equations:
10x+y+z=603
8x+2y+z=603
20x-10y-2z=-6
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
x=50 y=100 z=3
10x+y+z=603
8x+2y+z=603
20x-10y-2z=-6
we can set the first two equal to each other
10x+y+z=8x+2y+z
subtract z
10x+y=8x+2y
2x=y
multiply first by 2
20x+2y+2z=1206
20x-10y-2z=-6
subtract third from new first
12y+4z=1212
divide by 4
3y+z=303
8x+2y+z=603
add 300
3y+z+300=603
8x+2y+z=3y+z+300
8x=y+300
2x=y
8x=2x+300
6x=300
x=50
2x=y
y=100
3y+z=303
300+z=303
z=3
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