A young artist had one jar of blue and one jar of yellow paint, each enough to cover 38 dm² of area. Using all of this paint, he painted a picture: a blue sky, green grass, and a yellow sun. He obtained the green color by mixing two parts of yellow paint with one part of blue. What area of his picture is painted with each color, if the area of the grass in the picture is 6 dm² more than the area of the sky?
Let the areas painted blue, green, and yellow be denoted as $B, G$, and $Y$ respectively. Since the green color is obtained by mixing two parts of yellow paint and one part of blue, the amount of yellow paint used to paint the green area $G$ is equivalent to the area of $2 / 3 G$, and the amount of blue paint is $-1 / 3$ G. Since all the blue paint was used up, $B+1 / 3 G=38$. Additionally, according to the condition, $B=G-6$.

Substituting the value of $B$, we get $G=33$, so $B=27$.

Since all the yellow paint was also used up, $Y+2 / 3 G=38$. Substituting $G=33$, we get $Y=16$.

## Answer

The area painted blue is 27 dm², the area painted green is 33 dm², and the area painted yellow is 16 dm².