Answer:
3 kilometers per hour
Explanation:
Please consider the complete question.
Hong swam 8 kilometers against the current in the same amount of time it took him to swim kilometers 16 with the current. The rate of the current was 1 kilometer per hour. How fast would Hong swim if there were no current?
Let x represent Hang's swimming rate.
Hang's rate upstream would be:
![x-1](https://img.qammunity.org/2021/formulas/mathematics/high-school/za13p2w62u8wfsrxjbrtwmjqqkdkmtpblv.png)
Hang's rate downstream would be:
![\text{Time}=\frac{\text{Distance}}{\text{Rate}}](https://img.qammunity.org/2021/formulas/mathematics/high-school/46s58vtv45pd9z8fxi0piga4wagk0zr5yx.png)
![\text{Time taken downstream}=(16)/(x+1)](https://img.qammunity.org/2021/formulas/mathematics/high-school/wi7sz7ffrfyn8ja5awnt5yjzhjk8m23f8f.png)
![\text{Time taken against current}=(8)/(x-1)](https://img.qammunity.org/2021/formulas/mathematics/high-school/je0amreob1hhb9wwdqn30f0u4x2cki967q.png)
Since downstream and upstream times are equal, so we can equate both equation as:
![(16)/(x+1)=(8)/(x-1)](https://img.qammunity.org/2021/formulas/mathematics/high-school/c7su41bsdfg8v30iag6ka8bvoxmkikuunh.png)
Cross multiply:
![16(x-1)=8(x+1)](https://img.qammunity.org/2021/formulas/mathematics/high-school/iatgcbff7tzhrk1px9gn2hhoibih52d92l.png)
![16x-16=8x+8](https://img.qammunity.org/2021/formulas/mathematics/high-school/yg7jfn18w34ikvt37ddsjgltoefob177dl.png)
![16x-8x-16=8x-8x+8](https://img.qammunity.org/2021/formulas/mathematics/high-school/s5a0lzbea1fkpq0xth78amdkya2xcacsna.png)
![8x-16=8](https://img.qammunity.org/2021/formulas/mathematics/high-school/6eo3sxbxpv1n136wut5kx1skrlf5o8eyh0.png)
![8x-16+16=8+16](https://img.qammunity.org/2021/formulas/mathematics/high-school/1rwsuaagp85wgzb0m1nktbh62rj03nq2iu.png)
![8x=24\\\\(8x)/(8)=(24)/(8)\\\\x=3](https://img.qammunity.org/2021/formulas/mathematics/high-school/6yg8pfeoso9cgghkqsqht996lo1jqy4f34.png)
Therefore, Hong would swim at a rate of 3 kilometers per hour, if there were no current.