Can someone please explain how to do this.

Well you can graph it by plotting 3 points then drawing a line through these 3 points. The graph will be a straight line . By plotting 3 points you can be more sure if you are right, because if you make a mistake then they might not make a straight line.

Put y = 0 in the equation and find the x coordinate:-

x + 5(0) = -20

x = -20

So we have one point to plot:- (-20,0)

Putting x = 0 we get 5y = -20 so y = -4

Second point:- (0, -4)

Putting x = 5 say we get 4 + 5y = -20 so y = -25/5 = -5

so our 3rd point is (5, -5)

jdoe0001 jdoe0001 · Answer 2 · 2018-09-14T22:36:18+02:00

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.

[tex]\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)}[/tex]

and then we plot those, and run a line through them, check the picture below.