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

[latex] \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. [/latex] Ответ: (1;7)

{(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)