let's say the numbers are "a" and "b", we know their sum is
[tex]\bf 7\frac{2}{8}\implies 7\frac{1}{4}\qquad thus\qquad a+b=7\frac{1}{4}\implies a+b=\cfrac{7\cdot 4+1}{4}\implies a+b=\cfrac{29}{4}[/tex]
we also know that their difference is
[tex]\bf a-b=2\frac{4}{8}\implies a-b=2\frac{1}{2}\implies a-b=\cfrac{2\cdot 2+1}{2}\implies a-b=\cfrac{5}{2} \\\\[-0.35em] ~\dotfill\\\\ a+b=\cfrac{15}{2}\implies \boxed{b}=\cfrac{15}{2}-a \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{if we do some substitution on the 2nd equation}}{a-b=\cfrac{5}{2}\implies a-\left( \boxed{\cfrac{29}{4}-a} \right)=\cfrac{5}{2}}\implies a-\cfrac{29}{4}+a=\cfrac{5}{2}[/tex]
[tex]\bf 2a-\cfrac{29}{4}=\cfrac{5}{2}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{4}}{4\left( 2a-\cfrac{29}{4} \right)=4\left( \cfrac{5}{2} \right)} \implies 8a-29=10 \\\\\\ 8a=39\implies a=\cfrac{39}{8}\implies \blacktriangleright a = 4\frac{7}{8} \blacktriangleleft \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \stackrel{\textit{since we know that}}{b=\cfrac{29}{4}-a}\implies b=\cfrac{29}{4}-\cfrac{39}{8}\implies b = \stackrel{\textit{using the LCD of 8}}{\cfrac{58-39}{8}} \\\\\\ b = \cfrac{19}{8}\implies \blacktriangleright b = 2\frac{3}{8} \blacktriangleleft[/tex]
