Question

Решите систему уравнений:
3( x + y ) + 1 = x + 4y,
7 - 2( x - y ) = x - 8y

Answer 1

$\displaystyle \left \{ {{3(x+y)+1=x+4y} \atop {7-2(x-y)=x-8y}} \right. \\\\\left \{ {{x=\frac{1}{2}y-\frac{1}{2} } \atop {7-2(x-y)=x-8y}} \right.\\ \\7-2(\frac{1}{2}y-\frac{1}{2}-y)=\frac{1}{2}y-\frac{1}{2}-8y\\ \\y=-1\\ \\x=\frac{1}{2}(-1)-\frac{1}{2}\\ \\x=-1\\ (x,y)=(-1,-1)$

Answer 2

x = 1/2y - 1/2
7-2(x-y) =X-8y

7-2(1/2y-1/2-y) = 1/2y -1/2 - 8y
y = -1
x = 1/2 * (-1) - 1/2
x = -1

Ответ: (-1;-1)