{х÷у×(х+у)=-4, х÷у+(х+у)=-3

Замена переменной
х÷у=u; x+y=v
Решаем систему
$\left \{ {{uv}=-4 \atop {u+v=-3}} \right. \\ \\ \left \{ {{u\cdot(-3-u)}=-4 \atop {v=-3-u}} \right. \\ \\ \left \{ {{u^2+3u-4=0 \atop {v=-3-u}} \right. \\ \\ \left \{ {{u_1=-4; u_2=1 \atop {v_1=1;v_2=-4}} \right.$
1) $\left \{ {{ \frac{x}{y} =-4} \atop {x+y=1}} \right. \\ \\ \left \{ {{x=-4y} \atop {-4y+y=1}} \right. \\ \\ \left \{ {{y=- \frac{1}{3} } \atop {x= \frac{4}{3} }} \right.$

2) $\left \{ {{ \frac{x}{y} =1} \atop {x+y=-4}} \right. \\ \\ \left \{ {{x=y} \atop {2x=-4}} \right. \\ \\ \left \{ {{y=-2} \atop {x=-2}} \right.$