Dividing the second equation by 5 we get:
[tex]\begin{gathered} \frac{5y}{5}=\frac{10x-15}{5}, \\ y=2x-3. \end{gathered}[/tex]
Substituting the last equation in the first one we get:
[tex]-5x+2x-3=-12.[/tex]
Adding like terms we get:
[tex]-3x-3=-12.[/tex]
Adding 3 to the above equation we get:
[tex]\begin{gathered} -3x-3+3=-12+3, \\ -3x=-9. \end{gathered}[/tex]
Dividing the above equation by -3 we get:
[tex]\begin{gathered} \frac{-3x}{-3}=\frac{-9}{-3}, \\ x=3. \end{gathered}[/tex]
Finally, substituting x=3 in y=2x-3 we get:
[tex]\begin{gathered} y=2\cdot3-3, \\ y=6-3. \\ y=3. \end{gathered}[/tex]
Answer: One solution (3,3).
