Решите способом подстановки систему уравнений 2x^2-y^2=14 3x+2y=5

$\left \{ {{2x^2-y^2=14} \atop {3x+2y=5}} \right.$

$\left \{ {{2x^2-y^2=14} \atop {2y=5-3x}} \right.$

$\left \{ {{2x^2-y^2=14} \atop {y=2.5-1.5x}} \right.$

$\left \{ {{y=2.5-1.5x} \atop {2x^2-(2.5-1.5x)^2-14=0}} \right.$

$\left \{ {{y=2.5-1.5x} \atop {2x^2-6.25-2.25x^2+7.5x-14=0}} \right.$

$\left \{ {{y=2.5-1.5x} \atop {-0.25x^2+7.5x-20.25=0}} \right.$

$\left \{ {{y=2.5-1.5x} \atop {x^2-30x+81=0}} \right.$

$D=(-30)^2-4*1*81=900-324=576$

$x_1= \frac{30+24}{2}=27$

$x_2= \frac{30-24}{2}=3$

$\left \{ {{x_1=27} \atop {y_1=2.5-1.5*27}} \right.$ или $\left \{ {{x_2=3} \atop {y_2=2.5-1.5*3}} \right.$

$\left \{ {{x_1=27} \atop {y_1=-38}} \right.$ или $\left \{ {{x_2=3} \atop {y_2=-2}} \right.$

Ответ: $(27;-38)$ ; $(3;-2)$