Question

Нужно решить систему (2x+4)^2=3y (4x+2)^2=3y

Answer 1

$\left \{ {{(2x+4)^2=3y,} \atop {(4x+2)^2=3y;}} \right. \left \{ {{(2x+4)^2=3y,} \atop {(4x+2)^2=(2x+4)^2;}} \right. \left \{ {{(2x+4)^2=3y,} \atop { \left [ {{4x+2=2x+4,} \atop {4x+2=-(2x+4);}} \right.}} \right. \left [ {{ \left \{ {{(2x+4)^2=3y,} \atop {2x=2;}} \right. } \atop { \left \{ {{(2x+4)^2=3y,} \atop {6x=-6;}} \right.}} \right. \\$
$\left [ {{ \left \{ {{x=1,} \atop {3y=36;}} \right. } \atop { \left \{ {{x=-1,} \atop {3y=4;}} \right.}} \right. \left [ {{ \left \{ {{x=1,} \atop {y=12;}} \right. } \atop { \left \{ {{x=-1,} \atop {y=1 \frac{1}{3} ;}} \right.}} \right. \\ (1;12), (-1;1 \frac{1}{3}).$

Answer 2

$\left \{ {{(2x+4)^2=3y} \atop {(4x+2)^2=3y}} \right.\\ \\ (2x+4)^2=(4x+2)^2\\ 4x^2+16x+16=16x^2+16x+4\\ 4x^2+16x+16-16x^2-16x-4=0\\ -12x^2 +12=0\\ -12x^2=-12 \ \ |:(-12)\\ x^2=1\\ \\ \left[\begin{array}{ccc}x_1=1\\x_2=-1\end{array}\right \\ y_1=\frac{(2x_1+4)^2}3=\frac{(2*1+4)^2}{3}=\frac{36}3=12\\ y_2=\frac{(2x_2+4)^2}3=\frac{(2*(-1)+4)^2}{3}=\frac{4}3\\$
ответ: (1; 12) или (-1; 4/3)