We will ahve the following:
[tex]2x\le5x=-15[/tex]
This problem has the solution.
[tex]\Rightarrow2x+15\le5x+15=0\Rightarrow1\le\frac{5x+15}{2x+15}=0[/tex]
After this point you can reduce further.
[tex]1\le\frac{5x+15}{2x+15}=0[/tex]
We would need a value of "x" that makes "5x + 15" = 0 [We cannot have that "2x + 15 = 0" since it would be undetermined] so:
[tex]\rightarrow5x+15=0\rightarrow5x=-15\rightarrow x=-3[/tex]
Now, when we replace this value:
[tex]2(-3)\le5(-3)=-15\Rightarrow-6\le-15=-15[/tex]
