ANSWER
![y=-3x-7](https://img.qammunity.org/2023/formulas/mathematics/college/h4er0k20sudn2um72oypn755m1y2ax9hz3.png)
Step-by-step explanation
We want to find the equation of the line passing through (-3, 2) and parallel to:
![6x+2y=1](https://img.qammunity.org/2023/formulas/mathematics/college/83t7pgsx2o3g64idn6yw3x66ohs1agwcjd.png)
First, put the equation in slope intercept form:
![\begin{gathered} 2y=-6x+1 \\ y=-3x+(1)/(2) \\ \text{where slope, m = -3} \\ in\text{tercept, b = }(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/cfqsh1iphbg1hyn7wdui7glsaeief7xmoj.png)
A line that is parallel to another line has the same slope as that line.
Therefore, the slope of the required line is -3.
Now, use the point-slope method to find the equation of the line:
![y-y1=m(x-x1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/w8833wq43gzv073odoa76ku9bunhdf4cj7.png)
where (x1, y1) is the point the line passes through
Therefore, the equation of the line is:
![\begin{gathered} y-2=-3(x-(-3)) \\ y-2=-3(x+3) \\ y-2=-3x-9 \\ \Rightarrow y=-3x-9+2_{} \\ y=-3x-7 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wsw6p8yupkccuhxqest2wk3u7k81w5098p.png)
That is the equation of the line.