Question

Реши систему уравнений:
{x2−y2=48
x−y=6

{x=
y=

Answer 1

Ответ:

{(x-y)(x+y)=48

x=6+y

6(6+y+y)=48

36+12y=48

12y=12

y=1

x=6+1=7

{x=7

y=1

Объяснение:

Лень мне

Answer 2

$\left \{ {{x^2-y^2=48} \atop {x-y=6}} \right. => \left \{ {{(x-y)(x+y)=48} \atop {x-y=6}} \right. => \left \{ {{6*(x+y)=48} \atop {x-y=6}} \right.=>\left \{ {{x+y=8} \atop {x-y=6}} \right.=>\left \{ {{x=8-y} \atop {x-y=6}} \right.=>\\\\\left \{ {{x=8-y} \atop {8-y-y=6}} \right.=>\left \{ {{x=8-y} \atop {8-2y=6}} \right.=>\left \{ {{x=8-y} \atop {2y=2}} \right.=>\left \{ {{x=8-y} \atop {y=1}} \right.=>\left \{ {{x=8-1} \atop {y=1}} \right.=>\left \{ {{x=7} \atop {y=1}} \right.$

Применяем формулу разности квадратов $a^2-b^2=(a-b)(a+b)$, затем решаем методом подстановки.

Ответ:

${ {{x=7} \atop {y=1}} \right.$.