Let x and y be the two numbers we are trying to find.
'One number is 9 more 3 times another' is equivalent to x=9+3y
'Their product is 9 more than 3 times their sum' is equivalent to xy=9+3(x+y)
We have two equations and two unknowns; therefore, we can solve the problem
![\begin{cases}x=9+3y \\ xy=9+3(x+y)\end{cases}](https://img.qammunity.org/2023/formulas/mathematics/high-school/sp9c8fbke22xw82tnemk55xronh2nw1hm1.png)
Solve it as shown below
![\begin{gathered} x=9+3y \\ \Rightarrow xy=(9+3y)y=9y+3y^2 \\ \text{and} \\ 9+3(x+y)=9+3(9+3y+y)=9+3(9+4y)=9+27+12y=36+12y \\ \Rightarrow9y+3y^2=36+12y \\ \Rightarrow3y^2-3y-36=0 \\ \Rightarrow y^2-y-12=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/gt4ovtqa68ocsemxgypu42c21um9ikwz12.png)
Solve the quadratic equation
![\begin{gathered} y^2-y-12=0 \\ \Leftrightarrow(y-4)(y+3)=0 \\ \Rightarrow y=4,y=-3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/kkkoc4mdgjxfgs24r47q1y3zsp5ghs3nce.png)
Finally,
![\begin{gathered} y=4 \\ \Rightarrow x=9+3(4)=21 \\ \Rightarrow(21,4) \\ y=-3 \\ \Rightarrow x=9+3(-3)=0 \\ \Rightarrow(0,-3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/pjinwacvjsrlylm4qieu62g2u1623oqtmu.png)
The solutions are (21,4) and (0,-3)