Let us assume the numbers are x and y.
The first part of the question can be written as
[tex]x+y=200\text{ ---------------(1)}[/tex]and the second part can be written as
[tex]x-y=28\text{ --------------(2)}[/tex]From equation 1, we can get a value for y as
[tex]y=200-x\text{ -------------(3)}[/tex]Substitute for y in equation 3 into equation 2:
[tex]x-(200-x)=28[/tex]Expanding and solving, we get
[tex]\begin{gathered} x-200+x=28 \\ 2x=200+28 \\ 2x=228 \\ \therefore \\ x=\frac{228}{2} \\ x=114 \end{gathered}[/tex]Next, we substitute for the value of x into equation 3:
[tex]\begin{gathered} y=200-114 \\ y=86 \end{gathered}[/tex]Therefore, the two numbers are 114 and 86
