Question

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

Answer 1

1)
$frac{1}{3}y+2=1\frac{1}{3}y=1-2\frac{1}{3}y=-1\y=-1*3\y=-3$

2)
$frac{1}{5}x+11=1-frac{3}{5}x\frac{4}{5}x=-10\x=frac{-10*5}{4}\x=-12.5$

3)
$8-frac{1}{4}z=1\frac{1}{4}z=7\z=28$

4)
$3-frac{5}{7}t=1-frac{3}{7}t\frac{5}{7}t-frac{3}{7}t=3-1\frac{2}{7}t=2\t=frac{7*2}{2}\t=7$

5)
$frac{1}{8}u-2=frac{5}{8}u+1\frac{5}{8}u-frac{1}{8}u=-2-1\frac{1}{2}u=-3\u=-6$