We are given the following equation:
![x+4y=-6](https://img.qammunity.org/2023/formulas/mathematics/college/hoadh0jbk6nm0phfs4v6z7w9ssdalyuo5z.png)
We can solve for "y" first by subtracting x to both sides:
![4y=-6-x](https://img.qammunity.org/2023/formulas/mathematics/college/2g1syssnbuzahh1vj8ni94vb45j1vs0o9k.png)
now we divide both sides by 4:
![y=-(6)/(4)-(x)/(4)](https://img.qammunity.org/2023/formulas/mathematics/college/731lqu1mddgmfyaoqthn45y14tbx85xbfs.png)
Simplifying:
![y=-(3)/(2)-(x)/(4)](https://img.qammunity.org/2023/formulas/mathematics/college/gd2ui153a5xjij3zi8fvr17gwtwucdgktn.png)
Since the equation is of the form:
![y=mx+b](https://img.qammunity.org/2023/formulas/mathematics/high-school/smsb8cbft03lwblmi49nf2l6jby2ofxzws.png)
This is the equation of a line. Therefore, to plot this line we only need to points. We find the points by giving values to "x".
For x = 0:
![\begin{gathered} y=-(3)/(2)-(0)(1)/(4) \\ y=-(3)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ytmzf2q256mder1n6mksweehxjtl5on178.png)
Therefore the point (0, -3/2) is part of the line.
For x = 1
![\begin{gathered} y=-(3)/(2)-(1)/(4) \\ y=-(7)/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4nuo1ed4f6tjwj1c2x86biidmuyjn5p9gv.png)
therefore, the point (1, -7/4) is part of the line. Now we plot this to points and join them with a line, like this: