Решить систему:а) х-у=1 1/х+1/у=5/6б) 4х+ху=6 3х-5ху=39

$\left \{ {{ x=1+y} \atop {1/x+1/y=5/6}} \right.$
$\frac{1}{1+y} + \frac{1}{y} - \frac{5}{6} =0$
$\frac{6y+6+6y-5y-5y^2}{6y(1+y)} =0$
$-5y^2+7y+6=0; y \neq 0;y \neq -1$
$5y^2-7y-6=0; y \neq 0;y \neq -1$
D=49+120=169
y1=2
y2=-3/5
x1=1+y=3
x2=1+y=2/5

$\left \{ {{4x+xy=6} \atop { 3x-5xy=39}} \right.$
$\left \{ {{20x+5xy=30} \atop { 3x-5xy=39}} \right.$
$\left \{ {{20x+5xy=30} \atop { 23x=69}} \right.$
23x=69
x=3
4x+xy=6; 12+3y=6; 3y=-6; y=-2