313. $x-2y+4=0$, $x+y-5=0$ and $y=0$.
Solution. $1^{0}$. Let's construct the figure (Fig. 158).

Construct the line $x-2 y+4=0: \quad y=0, x=-4, A(-4 ; 0) ; x=0$, $y=2, \quad B(0 ; 2)$. Construct the line $x+y-5=0: y=0, x=5, C(5 ; 0)$; $x=0, y=5, D(0 ; 5)$.

$2^{6}$. Find the intersection point of the lines by solving the system

$$
\left\{\begin{array}{l}
x-2 y+4=0 \\
x+y-5=0
\end{array}\right.
$$

From this, $x=2, y=3$, i.e., $M(2 ; 3)$.

To calculate the desired area, we will divide the triangle $A M C$ into two triangles $A M N$ and $N M C$, since when $x$ changes from $A$ to $N$, the area is bounded by the line $x-2 y+4=0$, and when $x$ changes from $N$ to $C$, it is bounded by the line $x+y-5=0$.

$3^{0}$. For triangle $A M N$ we have $x-2 y+4=0 ; y=\frac{1}{2} x+2 ; f(x)=$ $=\frac{1}{2} x+2 ; a=-4 ; b=2$. For triangle $NMC$ we get $x+y-5=0$; $y=-x+5 ; f(x)=-x+5 ; a=2 ; b=5$.

$4^{0}$. Calculate the area of each of these triangles:

$$
\begin{aligned}
& S_{A M N}=\int_{-4}^{2}\left(\frac{1}{2} x+2\right) d x=\left.\left(\frac{x^{2}}{4}+2 x\right)\right|_{-4} ^{2}=9 \text { (sq. units) } \\
& S_{N M C}=\int_{2}^{5}(-x+5) d x=\left.\left(-\frac{x^{2}}{2}+5 x\right)\right|_{2} ^{5}=4.5 \text { (sq. units) }.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-324.jpg?height=294&width=437&top_left_y=276&top_left_x=197)

Fig. 158

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-324.jpg?height=354&width=387&top_left_y=690&top_left_x=212)

Fig. 160

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-324.jpg?height=392&width=357&top_left_y=187&top_left_x=704)

Fig. 159

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-324.jpg?height=298&width=373&top_left_y=747&top_left_x=661)

Fig. 161

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-324.jpg?height=325&width=419&top_left_y=1147&top_left_x=123)

Fig. 162

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-324.jpg?height=282&width=512&top_left_y=1182&top_left_x=613)

Fig. 163

Therefore,

$$
S=S_{A M N}+S_{N M C}=9+4.5=13.5 \text { (sq. units) }.
$$

Verification: $S \Delta_{A M C}=\frac{1}{2} A C \cdot N M=\frac{1}{2} \cdot 9 \cdot 3=13.5$ (sq. units).