Question

Решите систему уравнений (4х+3)^2=7у (3х+4)^2=7у

Answer 1

$\left \{ {{(4x+3)^{2}=7y} \atop {(3x+4)^{2}=7y}} \right. \\ \left \{ {{(4x+3)^{2}=(3x+4)^{2}} \atop {(3x+4)^{2}=7y}} \right. \\ \left \{ {{4x+3=3x+4} \atop {(3x+4)^{2}=7y}} \right. \\ \left \{ {{x=1} \atop {(3*1+4)^{2}=7y}} \right. \\ \left \{ {{x=1} \atop {49=7y}} \right. \\ \left \{ {{y=7} \atop {x=1}} \right.$

Ответ: (1;7)

Answer 2

{(4x+3)²=7y
{(3x+4)²=7y     |*(-1)

{(4x+3)²=7y
{-(3x+4)²=-7y
Метод сложения:
{(4x+3)²-(3x+4)²=7y-7y
{y=(4x+3)² /7

Решим первое уравнение:
(4x+3)²-(3x+4)²=7y-7y
16x²+24x+9-(9x²+24x+16)=0
16x²+24x+9-9x²-24x-16=0
7x²-7=0
x²=1
x=1 или x=-1

{x=1
{y=-7

{x=-1
{y=1/7

Ответ:(1;7), (-1;1/7)