Найти корень уравнения: 1/3y+2=1; 1/5x+11=1-3/5x; 8-1/4z=1; 3-5/7t=1-3/7t; 1/8u-2=5/8u+1; 2/5z-7=3

1) [latex]\frac{1}{3}y+2=1\\\frac{1}{3}y=1-2\\\frac{1}{3}y=-1\\y=-1*3\\y=-3[/latex] 2) [latex]\frac{1}{5}x+11=1-\frac{3}{5}x\\\frac{4}{5}x=-10\\x=\frac{-10*5}{4}\\x=-12.5[/latex] 3) [latex]8-\frac{1}{4}z=1\\\frac{1}{4}z=7\\z=28[/latex] 4) [latex]3-\frac{5}{7}t=1-\frac{3}{7}t\\\frac{5}{7}t-\frac{3}{7}t=3-1\\\frac{2}{7}t=2\\t=\frac{7*2}{2}\\t=7[/latex] 5) [latex]\frac{1}{8}u-2=\frac{5}{8}u+1\\\frac{5}{8}u-\frac{1}{8}u=-2-1\\\frac{1}{2}u=-3\\u=-6[/latex]