To do this, this equation must be taken to the form
[tex]\begin{gathered} y=mx+b \\ \text{Where} \\ m\colon\text{Slope oth the line} \\ b\colon\text{ intercept with the y-axis} \end{gathered}[/tex]So,
[tex]\begin{gathered} 2x-5y=-15 \\ \text{ Substract 2x from both sides of the equation} \\ 2x-5y-2x=-15-2x \\ -5y=-15-2x \\ \text{Divide by -5 from both sides of the equation} \\ \frac{-5y}{-5}=\frac{-15}{-5}-\frac{2}{-5}x \\ y=3+\frac{2}{5}x \\ y=\frac{2}{5}x+3 \end{gathered}[/tex]Then a positive slope of 2/5 tells you that for every 2 units on the y-axis there are 5 units on the x-axis. And 3 tells you that the line intercepts the y-axis at 3.
