Решите системы уравнений: 1) {7x+3y=12, {15x+y=4. 2) {x-y=10, {x*y=144

$\left \{ {{7x+3y=12} \atop {15x+y=4}} \right. \left \{ {{7x+3y=12} \atop {y=4-15x}} \right. 7x+3(4-15x)=12 7x+12-15x=12 -8x=0 x=0 y=4$
Ответ: (0;4)

$\left \{ {{x-y=10} \atop {x*y=144}} \right. \left \{ {{x=10+y} \atop {y(10+y)=144}} \right. y^{2} +10y-144=0 D=676 y_{1}= \frac{-10+26}{2} = 8 y_{2}= \frac{-10-26}{2}= -18 x_{1}= 18 x_{2}= -8$

Ответ: (18; 8) (-8;-18)