作业帮求下列一元一次方程的解[2(z+1)]/4-（7z-3）/20=-1/55-(1-3x)=35(x+8=6(2x-7)50﹪（y-1）=40﹪(y+1)+0.11/2(y-2)+1/6(y-2)-1/3(y-2)=1
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![求下列一元一次方程的解[2(z+1)]/4-（7z-3）/20=-1/55-(1-3x)=35(x+8=6(2x-7)50﹪（y-1）=40﹪(y+1)+0.11/2(y-2)+1/6(y-2)-1/3(y-2)=1](/uploads/image/z/11548607-23-7.jpg?t=%E6%B1%82%E4%B8%8B%E5%88%97%E4%B8%80%E5%85%83%E4%B8%80%E6%AC%A1%E6%96%B9%E7%A8%8B%E7%9A%84%E8%A7%A3%5B2%28z%2B1%29%5D%2F4-%EF%BC%887z-3%EF%BC%89%2F20%3D-1%2F55-%281-3x%29%3D35%28x%2B8%3D6%282x-7%2950%EF%B9%AA%EF%BC%88y-1%EF%BC%89%3D40%EF%B9%AA%28y%2B1%29%2B0.11%2F2%28y-2%29%2B1%2F6%28y-2%29-1%2F3%28y-2%29%3D1)
求下列一元一次方程的解[2(z+1)]/4-（7z-3）/20=-1/55-(1-3x)=35(x+8=6(2x-7)50﹪（y-1）=40﹪(y+1)+0.11/2(y-2)+1/6(y-2)-1/3(y-2)=1
求下列一元一次方程的解
[2(z+1)]/4-（7z-3）/20=-1/5
5-(1-3x)=3
5(x+8=6(2x-7)
50﹪（y-1）=40﹪(y+1)+0.1
1/2(y-2)+1/6(y-2)-1/3(y-2)=1
求下列一元一次方程的解[2(z+1)]/4-（7z-3）/20=-1/55-(1-3x)=35(x+8=6(2x-7)50﹪（y-1）=40﹪(y+1)+0.11/2(y-2)+1/6(y-2)-1/3(y-2)=1
z=-11/3
x=-1/3
x=82/7
y=10
y=5