1. **Stating the problem:** Determine the system of linear equations (SPLDV) from the given graphs labeled 4 and 5. 2. **Understanding SPLDV:** SPLDV means Sistem Persamaan Linear Dua Variabel, or system of two linear equations with two variables $x$ and $y$. 3. **For graph 4:** - The graph shows two lines intersecting near $(3,2)$. - One line goes downward from $(0,4)$ to $(6,0)$, so its slope is $m=\frac{0-4}{6-0}=-\frac{4}{6}=-\frac{2}{3}$. - Equation of this line using point-slope form: $$y-4 = -\frac{2}{3}(x-0) \Rightarrow y = -\frac{2}{3}x + 4$$ - The other line goes upward from $(0,-2)$ and passes through $(3,2)$. - Slope is $m=\frac{2-(-2)}{3-0}=\frac{4}{3}$. - Equation: $$y + 2 = \frac{4}{3}x \Rightarrow y = \frac{4}{3}x - 2$$ 4. **SPLDV from graph 4:** $$\begin{cases} y = -\frac{2}{3}x + 4 \\ y = \frac{4}{3}x - 2 \end{cases}$$ 5. **For graph 5:** - The graph shows a line ascending from $(0,0)$ to near $(10,6)$. - Slope is $m=\frac{6-0}{10-0}=\frac{6}{10}=\frac{3}{5}$. - Equation: $$y = \frac{3}{5}x$$ - Vertical boundaries at $x=2$ and $x=4$ and horizontal boundary at $y=4$ form a rectangle. 6. **SPLDV for graph 5:** - The line equation: $$y = \frac{3}{5}x$$ - The vertical boundaries: $$x = 2$$ $$x = 4$$ - The horizontal boundary: $$y = 4$$ 7. **Summary:** - Graph 4 SPLDV: $$\begin{cases} y = -\frac{2}{3}x + 4 \\ y = \frac{4}{3}x - 2 \end{cases}$$ - Graph 5 SPLDV: $$\begin{cases} y = \frac{3}{5}x \\ x = 2 \\ x = 4 \\ y = 4 \end{cases}$$