诬陷罪:2x+Y+3z=23,x+4y+5z=36,求X+2Y+3z~~~~
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要过程~~!
2x+Y+3z=23...(1)
x+4y+5z=36...(2)
(2)*2-(1)
7Y+7Z=49
Y+Z=7
x+4y+5z=36=X+2Y+3Z+2(Y+Z)
X+2Y+3Z=36-2*7=22
2x+Y+3z=23...... A
x+4y+5z=36...... B
A-2B:-7y-7z=-49
y+z=7
y=7-z
4A-B:7x+7z=56
x+z=8
x=8-z
X+2Y+3z=(8-z)+2(7-z)+3z=22
2x+y+3z=23,x+4y+5z=36
4x+2y+6z=46,3x+12y+15z=108
两式相加:
7x+14y+21z=154
x+2y+3z=22
令X+2Y+3z=t(2x+Y+3z)+s(x+4y+5z),解粗s,t来就成.
2x+Y+3z=23,x+4y+5z=36,求X+2Y+3z~~~~
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2x+5y+4z=0,3x+y-7z=0,则x+y-z=?
2x+5y+4z=0,3x+y-7z=0,则x+y-z=?
13x+5y+9z=12.7 2x+4y+3z=4.7 x+y+z=?
x =2y,3y=4z,则x:y:z =?:?:?
已知2x+Y+3z=23,x=4y+5z=36,求X+2Y+3Z~~~~
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