1. For real numbers $a$ and $b$, it holds that $\frac{4 a}{a+2 b}-\frac{5 b}{2 a+b}=1$. What value can $\frac{a-2 b}{4 a+5 b}$ take?
IV/1. First, we observe that the expressions $a+2b$ and $2a+b$ cannot be equal to 0, so the numbers $a$ and $b$ cannot both be equal to 0. From the equation $\frac{4a}{a+2b} - \frac{5b}{2a+b} = 1$, we eliminate the fractions and get $8a^2 + 4ab - 5ab - 10b^2 = 2a^2 + 5ab + 2b^2$, which can be rewritten as $6(a+b)(a-2b) = 0$. If $a = -b$, the value of the expression $\frac{a-2b}{4a+5b}$ is $\frac{-3b}{b} = -3$, since $b$ must be different from 0 (since $a$ and $b$ cannot both be 0). If $a = 2b$, the value of the expression $\frac{a-2b}{4a+5b}$ is 0, as the denominator is non-zero again. Therefore, the value of the expression can be 0 or -3.

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-16.jpg?height=71&width=1636&top_left_y=1158&top_left_x=221)

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-16.jpg?height=57&width=1631&top_left_y=1214&top_left_x=224)

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-16.jpg?height=52&width=1633&top_left_y=1262&top_left_x=220)

(If there is no comment that the numbers $a$ and $b$ cannot both be 0, award only 1 point.)

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-16.jpg?height=51&width=1633&top_left_y=1365&top_left_x=220)

(If there is no comment that the numbers $a$ and $b$ cannot both be 0, award only 1 point.)

Conclusion that the value of the expression can be 0 or -3

1 point