Let's call one of the numbers as x and the other as y.
From the first sentence
"One number is 9 more than 3 times another."
From this sentence, we know that our first number(x) plus 9 is equal to 3 times the other(y).
![x+9=3y](https://img.qammunity.org/2023/formulas/mathematics/college/93j6kuu50edil1tsecn0ik6s6cn410qloa.png)
From the other sentence
"Their product is 9 more than 3 times their sum"
We get the following equation
![xy+9=3(x+y)](https://img.qammunity.org/2023/formulas/mathematics/college/6c8ex6ed2d25wfvamkdcx3doi9esuexu58.png)
Now we have two equations for two variables
![\begin{cases}x+9=3y \\ xy+9=3(x+y)\end{cases}](https://img.qammunity.org/2023/formulas/mathematics/college/ymgs54kg9y2a7sc0r45kvg9glsu9tgd7h5.png)
Expanding the parentheses, we have
![\begin{cases}x+9=3y \\ xy+9=3x+3y\end{cases}](https://img.qammunity.org/2023/formulas/mathematics/college/wdgo1ux0cghushk27muvq0fbh1htiie7cs.png)
Using our value for 3y in the first equation in the second equation, we get a new equation
![\begin{gathered} xy+9=3x+3y \\ xy+9=3x+(x+9) \\ xy+9=3x+x+9 \\ xy+9=4x+9 \\ xy=4x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/dpnyj1u9eas6s667fg8hdpnrgh3civ0kp9.png)
This last equation have an immediate solution, x = 0. If x is not 0, we can divide both sides by x.
![(xy)/(x)=(4x)/(x)\Rightarrow y=4](https://img.qammunity.org/2023/formulas/mathematics/college/2h5anxxzx1l1jtp83aje6no27juxpq7q4j.png)
Using this value for y, we can evaluate any of the equations to find its corresponding x, and we can do the same for x = 0 to find the corresponding value for y.
![\begin{gathered} y=4\Rightarrow x+9=3\cdot4\Rightarrow x=3 \\ x=0\Rightarrow0+9=3y\Rightarrow y=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/divygqkn0zfmjm34q6vth2vvhr8zd7uim7.png)
This means, our points are (0, 3) and (3, 4).