1. **Problem Statement:** Given the functions $f$ and $g$ with the relation $$2 = \frac{x + (f \circ g)(x)}{f(x) + g(x)}$$ and the limit $$\lim_{x \to +\infty} 2 = \lim_{x \to +\infty} \frac{x + f(g(x))}{f(x) + g(x)},$$ find the value of $-a$. 2. **Understanding the problem:** We want to find $-a$ where $a$ is related to the functions $f$ and $g$ and their behavior as $x$ approaches infinity. 3. **Key points and assumptions:** - $f$ and $g$ intersect at the origin. - $f$ is increasing and passes through $(2a, y)$. - $g$ is decreasing. - The limit expression involves composition $f(g(x))$. 4. **Rewrite the limit expression:** $$2 = \lim_{x \to +\infty} \frac{x + f(g(x))}{f(x) + g(x)}$$ 5. **Analyze behavior as $x \to +\infty$:** - Since $f$ is increasing and $g$ is decreasing, as $x \to +\infty$, $g(x)$ likely tends to $-\infty$. - Then $f(g(x)) = f(\text{large negative})$. 6. **Assuming linearity for simplicity:** Let $f(x) = m_f x$ and $g(x) = m_g x$ where $m_f > 0$ and $m_g < 0$. 7. **Substitute into the limit:** $$\lim_{x \to +\infty} \frac{x + f(g(x))}{f(x) + g(x)} = \lim_{x \to +\infty} \frac{x + m_f (m_g x)}{m_f x + m_g x} = \lim_{x \to +\infty} \frac{x + m_f m_g x}{x(m_f + m_g)} = \frac{1 + m_f m_g}{m_f + m_g}$$ 8. **Set equal to 2:** $$2 = \frac{1 + m_f m_g}{m_f + m_g}$$ 9. **Solve for $m_f$ and $m_g$:** Multiply both sides by denominator: $$2(m_f + m_g) = 1 + m_f m_g$$ $$2 m_f + 2 m_g = 1 + m_f m_g$$ 10. **Rearranged:** $$m_f m_g - 2 m_f - 2 m_g + 1 = 0$$ 11. **Recall $f(2a) = y$ and $f$ passes through $(2a, y)$:** Since $f(x) = m_f x$, point $(2a, y)$ means $y = m_f (2a)$. 12. **From the graph, $-2$ is marked on x-axis, likely related to $g$ or $a$:** Assuming $2a = -2$ gives $a = -1$. 13. **Therefore, $-a = 1$**. **Final answer:** $$\boxed{1}$$