It is given in the question that,
At most, Keith can spend $90 on sandwiches and chips for a company lunch. He already bought chips for $15 and will buy sandwiches that cost $6 each. The inequality
![15 + 6s \leq 90](https://img.qammunity.org/2019/formulas/mathematics/high-school/tsgx2cpwifsgfqssgwcycti63zdm3gsmik.png)
represents the described cost, where s is the maximum number of sandwiches that could be bought.
Now we have to solve for s, and for that, we first subtract 15 to both sides, that is
![6s\leq 75](https://img.qammunity.org/2019/formulas/mathematics/high-school/i7lx8zgccbhm4s9aip7smzsxa1ty325e4l.png)
Dividing both sides by 6
![s\leq 12.5](https://img.qammunity.org/2019/formulas/mathematics/high-school/p34f8iy3eq49ka8m92j9cm6cf5j55jszuw.png)
So the maximum number of sandwiches Keith can buy is 12.