Let x be the first number and y be the second number, since the difference between 3 times x and y is 15, and the sum of 4 times x and y is 13, we can set the following system of equations:
[tex]\begin{gathered} 3x-y=15, \\ 4x+y=13. \end{gathered}[/tex]Adding both equations we get:
[tex]3x-y+4x+y=15+13.[/tex]Simplifying the above equation we get:
[tex]7x=28.[/tex]Dividing the above equation by 7 we get:
[tex]\begin{gathered} \frac{7x}{7}=\frac{28}{7}, \\ x=4. \end{gathered}[/tex]Now, substituting x=4 in 4x+y=13 we get:
[tex]\begin{gathered} 4\cdot4+y=13, \\ 16+y=13. \end{gathered}[/tex]Subtracting 16 from the above equation we get:
[tex]\begin{gathered} 16+y-16=13-16, \\ y=-3. \end{gathered}[/tex]Answer: The first number is 4 and the second is -3.
