Let x be the cost of one corsage, and y be the cost of one boutonniere, then we can set the following system of equations:
![\begin{gathered} 14x+10y=490, \\ 13x+11y=503. \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mqrolcfouifypc8ljvf71k6tkxncyq1ey8.png)
Solving the second equation for y, we get:
![\begin{gathered} 11y=503-13x, \\ y=(503)/(11)-(13)/(11)x\text{.} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qb63wtfghis0wrqnzhjy3u40d84kc7qxvb.png)
Substituting the above equation in the first one we get:
![14x+10((503)/(11)-(13)/(11)x)=490.](https://img.qammunity.org/2023/formulas/mathematics/college/76okkgv4axuro124mxxa9vax7w7b43emrp.png)
Solving for x, we get:
![\begin{gathered} 14x+(5030)/(11)-(130)/(11)x=490, \\ (24)/(11)x=490-(5030)/(11)=(360)/(11), \\ 24x=360, \\ x=15. \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/k6rtvr3wwo9xpoxpefzne4i5vr3vv3pmmj.png)
Substituting x=15 in the equation that we solved for y we get:
![y=(503)/(11)-(13*15)/(11)=28.](https://img.qammunity.org/2023/formulas/mathematics/college/zil5dzkt7djuwebcw4jsozjnqnbijpybv5.png)
Answer: A corsage sells for $15, and a boutonniere sells for $28.