so, we can firstly, solve for "y" on that equation, and then pick any random "x" values to get a "y", and therefore get a point, and we do that a few times, and then plot a line through those collinear points, since the graph of that equation, being a LINEar equation, is just a line, anyow, we only need two points to graph a line, but let's get 3 anyway.
we'll use say x = 0, x = 5, x = 10.
![\bf x+5y=-20\implies 5y=-20-x\implies y=\cfrac{-20-x}{5} \\\\\\ \stackrel{\textit{distributing the denominator}}{y=-\cfrac{20}{5}-\cfrac{x}{5}}\implies y=-4-\cfrac{x}{5} \\\\[-0.35em] ~\dotfill\\\\ x=0~\hspace{5em}y=-4-\cfrac{0}{5}\implies y=-4~\hfill \boxed{(0,-4)} \\\\[-0.35em] ~\dotfill\\\\ x=5~\hspace{5em}y=-4-\cfrac{5}{5}\implies y=-5~\hfill \boxed{(5,-5)} \\\\[-0.35em] ~\dotfill\\\\ x=10~\hspace{5em}y=-4-\cfrac{10}{5}\implies y=-6~\hfill \boxed{(10,-6)}](https://tex.z-dn.net/?f=%5Cbf%20x%2B5y%3D-20%5Cimplies%205y%3D-20-x%5Cimplies%20y%3D%5Ccfrac%7B-20-x%7D%7B5%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7B%5Ctextit%7Bdistributing%20the%20denominator%7D%7D%7By%3D-%5Ccfrac%7B20%7D%7B5%7D-%5Ccfrac%7Bx%7D%7B5%7D%7D%5Cimplies%20y%3D-4-%5Ccfrac%7Bx%7D%7B5%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0Ax%3D0~%5Chspace%7B5em%7Dy%3D-4-%5Ccfrac%7B0%7D%7B5%7D%5Cimplies%20y%3D-4~%5Chfill%20%5Cboxed%7B%280%2C-4%29%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0Ax%3D5~%5Chspace%7B5em%7Dy%3D-4-%5Ccfrac%7B5%7D%7B5%7D%5Cimplies%20y%3D-5~%5Chfill%20%5Cboxed%7B%285%2C-5%29%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0Ax%3D10~%5Chspace%7B5em%7Dy%3D-4-%5Ccfrac%7B10%7D%7B5%7D%5Cimplies%20y%3D-6~%5Chfill%20%5Cboxed%7B%2810%2C-6%29%7D)
and then we plot those, and run a line through them, check the picture below.