We will ahve the following:
![2x\le5x=-15](https://img.qammunity.org/2023/formulas/mathematics/college/ld603zwry3fyided0ey1x3wd2pmflpuy7u.png)
This problem has the solution.
![\Rightarrow2x+15\le5x+15=0\Rightarrow1\le(5x+15)/(2x+15)=0](https://img.qammunity.org/2023/formulas/mathematics/college/vc734ym7cidw628e7ncqynf4qik5btk86w.png)
After this point you can reduce further.
![1\le(5x+15)/(2x+15)=0](https://img.qammunity.org/2023/formulas/mathematics/college/bgnpj07fpjo31ip1a4bjhjwb44djkcrwmn.png)
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:
![\rightarrow5x+15=0\rightarrow5x=-15\rightarrow x=-3](https://img.qammunity.org/2023/formulas/mathematics/college/m75de0q6pym0povs88jjrr89odryqyqr14.png)
Now, when we replace this value:
![2(-3)\le5(-3)=-15\Rightarrow-6\le-15=-15](https://img.qammunity.org/2023/formulas/mathematics/college/2u9sood7q1fwtyta6tjztbcggl4gsjmibd.png)