The chef will end up with 300 mL of a dressing that is 9% vinegar. Let's calculate how much vinegar will the dressing have:
Therefore,
![v=(300*9)/(100)\rightarrow27](https://img.qammunity.org/2023/formulas/mathematics/college/rhzvxmi2xdcp45zol2xhzvcu193cwtgdzz.png)
He'll have 27mL of vinegar.
Now, we know that x mL of the 5% dressing plus y mL of the 15% one will give the chef a total of 27mL of vinegar. Therefore,
![\begin{gathered} x((5)/(100))+y((15)/(100))27 \\ \\ \rightarrow0.05x+0.15y=27 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6ogs5g9ro8izyvbjap7krz4kynomidq2aj.png)
And that x mL of the 5% dressing plus y mL of the 15% one will give the chef a total of 300mL of dressing. Therefore,
![x+y=300](https://img.qammunity.org/2023/formulas/mathematics/college/y4tmmudjtr4zludipik4xvu6ufjvwr9mmo.png)
We have a system of equations:
![\mleft\{\begin{aligned}0.05x+0.15y=27 \\ x+y=300\end{aligned}\mright.](https://img.qammunity.org/2023/formulas/mathematics/college/qmkdo5krsr8lfol4haucpmdlnps0nky878.png)
Let's clear y from equation 2 and replace in equation 1:
![\begin{gathered} x+y=300\rightarrow y=300-x \\ \\ 0.05x+0.15y=27\rightarrow0.05x+0.15(300-x)=27 \\ \rightarrow0.05x+45-0.15x=27\rightarrow-0.10x=-18\rightarrow x=(-18)/(-0.10) \\ \\ \rightarrow x=180 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/daqc4xihbiyuq8thcld4wsptjfhkgli715.png)
Thereby,
![\begin{gathered} y=300-x\rightarrow y=300-180 \\ \\ \rightarrow y=120 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/x4za3fj4wpos4jcyw6gc7cq4jfvtc6659w.png)
The chef would have to use 180mL of the dressing that's 5% vinegar, and 120mL of the one that's 15% vinegar.