解三元一次方程组{x/7=y/10=z/5,x+2y+3z=84
人气:484 ℃ 时间:2020-01-28 01:47:59
解答
∵x/7=y/10=z/5
∴x=7z/5
y=2z
又∵x+2y+3z=84
∴(7z/5)+2(2z)+3z=84
化简可得(7/5)z+7z=84
解得z=10
∴x=7z/5=(7*10)/5=14
y=2z=2*10=20
∴x=14,y=20 ,z=10
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