312. $x y=6$ and $x+y-7=0$.
Solution. Let's draw the figure whose area we need to determine (Fig. 157). We will find the abscissas of the points of intersection of the equilateral hyperbola and the line by solving the system of equations $\left\{\begin{array}{l}x y=6, \\ x+y-7=0 .\end{array}\right.$

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-323.jpg?height=332&width=542&top_left_y=184&top_left_x=354)

Fig. 157

We have $x(7-x)=6 ; 7 x-x^{2}=6 ; x^{2}-7 x+6=0$, from which $x_{1}=6, x_{2}=1$. Thus, $a=1, b=6$.

Therefore, to find the required area, we need to subtract the area of the curvilinear trapezoid $a A C B b$ from the area of the trapezoid $a A B b$. We find

$$
\begin{gathered}
S_{a A B b}=\int_{1}^{6}(7-x) d x=\left.\left(7 x-\frac{x^{2}}{2}\right)\right|_{1} ^{6}=\left(7 \cdot 6-\frac{36}{2}\right)-\left(7 \cdot 1-\frac{1}{2}\right)= \\
=17.5 \text { (sq. units); } \quad S_{a A C B b}=\int_{1}^{6} \frac{6 d x}{x}=\left.6 \ln x\right|_{1} ^{6}=6 \ln 6 \text { (sq. units) }
\end{gathered}
$$

i.e., $S=(17.5-6 \ln 6)$ sq. units.