6.073. $\left\{\begin{array}{l}y^{2}-x y=-12, \\ x^{2}-x y=28\end{array}\right.$
## Solution.

From the condition, we have $\left\{\begin{array}{l}y(y-x)=-12, \\ x(x-y)=28 .\end{array}\right.$

Dividing the first equation by the second, we get

$$
\left\{\begin{array} { l } 
{ \frac { y ( x - y ) } { x ( x - y ) } = \frac { 12 } { 28 } , } \\
{ x ( x - y ) = 28 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \frac { y } { x } = \frac { 3 } { 7 } , } \\
{ x ( x - y ) = 28 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y=\frac{3 x}{7} \\
x(x-y)=28
\end{array}\right.\right.\right.
$$

Substituting $y=\frac{3 x}{7}$ from the first equation of the system into the second, we get

$$
\left\{\begin{array} { l } 
{ y = \frac { 3 x } { 7 } } \\
{ x ( x - \frac { 3 x } { 7 } ) = 28 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y=\frac{3 x}{7} \\
x^{2}=49
\end{array}\right.\right.
$$

From here,

1) $\left\{\begin{array}{l}y=\frac{3 x}{7} \\ x=-7\end{array}, \Leftrightarrow\left\{\begin{array}{l}y_{1}=-3 \\ x_{1}=-7\end{array}\right.\right.$
or
2) $\left\{\begin{array}{l}y=\frac{3 x}{7} \\ x=7\end{array}, \Leftrightarrow\left\{\begin{array}{l}y_{2}=3, \\ x_{2}=7 .\end{array}\right.\right.$

Answer: $(-7 ;-3)(7 ; 3)$.