7.3. A young artist had one jar of blue and one jar of yellow paint, each enough to cover 38 dm ${ }^{2}$ 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 and one part of blue. What area on his picture is painted with each color, if the area of the grass in the picture is 6 dm $^{2}$ more than the area of the sky
Answer: Blue is shaded 27 dm $^{2}$, green - 33 dm $^{2}$, and yellow - 16 dm $^{2}$.

Solution. First method. Let the areas shaded blue (blue), green (green), and yellow (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 for shading the area $G$ is equivalent to the area $\frac{2}{3} G$, and the blue paint is $-\frac{1}{3} G$. Considering that all the blue paint was used, we can set up the equation: $B+\frac{1}{3} G=38$. Additionally, according to the problem, $G$ is 6 dm $^{2}$ more than $B$, that is, $B=G-6$. Substituting the value of $B$ into the equation, we get $G=33$, so $B=27$.

Since all the yellow paint was also used, $Y+\frac{2}{3} G=38$. Substituting the value $G=33$ into this equation, we get $Y=16$.

Second method. Let $x$ be one part used to obtain the green paint. Then the yellow paint covers ( $38-2 x$ ) dm², the green $-3 x$ dm $^{2}$, and the blue $-(38-x)$ dm ${ }^{2}$. Since according to the problem, the green paint covers 6 dm $^{2}$ more than the blue, we have $3 x-6=38-x$. From this, $x=11$, so the yellow paint covers $38-2 \cdot 11=16$ dm ${ }^{2}$, the green $3 \cdot 11=33$ dm $^{2}$, and the blue $38-11=27$ dm $^{2}$.

## Grading Criteria:

+ Correct answer and fully justified solution provided

$\pm$ Correct reasoning provided, all three areas found, but an arithmetic error was made

$\pm$ Correct reasoning provided, but only two out of three areas found

干 Correct equation (system of equations) set up, but not solved

- Only the answer provided
- Problem not solved or solved incorrectly