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:
![\begin{gathered} 3x-y=15, \\ 4x+y=13. \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ewqowzi7kp965pmm5npezkshpdqeqz050f.png)
Adding both equations we get:
![3x-y+4x+y=15+13.](https://img.qammunity.org/2023/formulas/mathematics/college/wna1sg00c67ztea74pod3os9huefjbed4f.png)
Simplifying the above equation we get:
![7x=28.](https://img.qammunity.org/2023/formulas/mathematics/college/w4kr6ilo7ugflskqpravmoq2jf3j2glx36.png)
Dividing the above equation by 7 we get:
![\begin{gathered} (7x)/(7)=(28)/(7), \\ x=4. \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/437j93r1z52ri9f5ctl5jwoylpf6i6tkeb.png)
Now, substituting x=4 in 4x+y=13 we get:
![\begin{gathered} 4\cdot4+y=13, \\ 16+y=13. \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/omrbt6ubk8a11dm4j0ezkyu1pmt0tavald.png)
Subtracting 16 from the above equation we get:
![\begin{gathered} 16+y-16=13-16, \\ y=-3. \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/13e2m83zz38x77u51i4fp61valhpln6o8l.png)
Answer: The first number is 4 and the second is -3.