三元一次方程!

1个回答

  • 甲乙丙每天能完成的工作量

    x + y = 5/12

    y + z = 4/15

    x + z = 7/20

    ==> x = 1/4 y = 1/6 z = 1/10

    甲乙丙每天的费用

    x + y = 1800 * 5/12 = 750

    y + z = 1500 * 4/15 = 400

    x + z = 1600 * 7/20 = 560

    ==> x = 455 y = 295 z = 105

    甲乙丙完成工程需要的费用

    甲 455 * 4 = 1820

    乙 295 * 6 = 1680

    丙 105 * 10 = 1050

    所以丙最合算,选择丙设单独购买甲、乙、丙一件依次需要x元、y元、z元.

    则:依题意,有:3x+7y+z=5.8、且4x+10y+z=6.3.

    由3x+7y+z=5.8、4x+10y+z=6.3两式相减,得:x+3y=0.5.

    由3x+7y+z=5.8、4x+10y+z=6.3两式相加,得:7x+17y+2z=12.1,

    ∴2(x+y+z)+3(x+3y)=12.1.

    将x+3y=0.5代入到2(x+y+z)+3(x+3y)=12.1中,得:2(x+y+z)+1.5=12.1,

    ∴2(x+y+z)=10.6,∴x+y+z=5.3(元).

    即:购买甲、乙、丙各一件需要5.3元.