The answer is a diagonal line that cuts through 0,0
Like in this website: https://www.symbolab.com/solver/step-by-step/5x%2B10y%3E0
My attempt: 5x + 10y(0) >0
5x>0
x>0
5x(0)+10y>0
10y>0
y>0
I can't see how you can draw that line.
Sorry for my ignorance.
On
Every point on the $x-y$ plane satisfies exactly one of these relations: $$5x+10y>0$$ $$5x+10y=0$$ $$5x+10y<0$$
The equality (the center one) is the line. The two inequalities are on one side of the line or the other.
So, you graph the line for the equality ($y$ intercept of $0$ and slope $-2$), and then choose a point not on the line to test whether it falls in the region you want. Let's choose $(1,1)$. Since
$$5(1) + 10(1) = 15 > 0,$$ $(1,1)$ is in the region you want, so you shade the side that contains $(1,1)$.
(When you graphed the regions using the tool at the link in your question, notice that the line is dashed. This is important, because it means that the region covered by the inequality does not include the line. If it was, say $5x+10y \geq 0$, then the line would be solid, and the line would be included in the solution.)
On
The inequality $5x+10y>0$ can be transformed to the equivalent $$y>-x/5.$$ In other words, the inequality is satisfied by all points $(x,y)$ whose $y$-coordinates are greater than $-x/5.$ To determine these points, first get those points satisfying the equation $y=-x/5.$ This is the equation of a line through the origin and therefore it divides the plane into two parts. Any point above this line for $x<0$ (i.e., whose ordinate crosses -- but is not on -- this line) lies in the region you want and determines it.
The set of points $(x,y)$ satisfying $5x+10y=0=x+2y$ is a line. this line cuts the plane into two parts. in one side, we have $$x+2y>0$$ and in the other, $$x+2y<0.$$
take for example, the point $A(x_A=1,y_A=0)$ which is not in the line. then $$x_A+2y_A=1>0.$$ the solution of $$x+2y>0$$ is the side containing A.