Question

Решите систему уравнений:
а) $\left \{ {{ x^{2} +4xy+4y ^{2}=7x+14y } \atop {2x+5y=15}} \right.$
б)$. \left \{ {{9y^{2}-6xy-5x+ x^{2}+15y=0 } \atop {3x-4y=20}} \right.$

Answer 1

$\left \{ {{x^2+4xy+4y^2=7x+14y} \atop {2x+5y=15}} \right. \left \{ {{(x+2y)^2=7(x+2y)} \atop {2x+5y=15}} \right. \left \{ {{x+2y=7} \atop {2x+5y=15}} \right. \left \{ {{x=7-2y} \atop {2(7-2y)+5y=15}} \right.$

$\left \{ {{x=7-2y} \atop {14-4y+5y=15}} \right. \left \{ {{x=7-2y} \atop {14+y=15}} \right. \left \{ {{x=7-2y} \atop {y=15-14}} \right. \left \{ {{x=7-2*1} \atop {y=1}} \right. \left \{ {{y=5} \atop {y=1}} \right.$


$\left \{ {{9y^2-6xy-5x+x^2+15y=0} \atop {3x-4y=20}} \right. \left \{ {{9y^2-6xy+x^2=5x-15y} \atop {3x-4y=20}} \right. \left \{ {{(3y-x)^2=-5(3y-x)} \atop {3x-4y=20}} \right.$

$\left \{ {{3y-x=-5} \atop {3x-4y=20}} \right. \left \{ {{-x=-5-3y} \atop {3x-4y=20}} \right. \left \{ {{x=5+3y} \atop {3x-4y=20}} \right. \left \{ {{x=5+3y} \atop {3(5+3y)-4y=20}} \right. \left \{ {{x=5+3y} \atop {15+9y-4y=20}} \right.$

$\left \{ {{x=5+3y} \atop {5y=20-15}} \right. \left \{ {{x=5+3y} \atop {5y=5}} \right. \left \{ {{x=5+3*1} \atop {y=1}} \right. \left \{ {{x=8} \atop {y=1}} \right.$