Question 44261
Let the two numbers be called "x" and "y".
The numbers differ by 5:
{{{x-y=5}}}.........................(i)

Rearange equation (i) to give x in terms of y:
{{{x=y+5}}}.........................(ii)



Their squares  differ by 55:
{{{x^2-y^2=55}}}....................(iii)
Substitute equation (ii) into equation (iii) for x:
{{{(y+5)^2-y^2=55}}}
{{{(y+5)(y+5)-y^2=55}}}
{{{y^2+10y+25-y^2=55}}}
{{{10y+25=55}}}
{{{10y=30}}}
{{{y=3}}}
substitute "y=3" into equation (ii) to give:
{{{x=3+5}}}
So x=8 and y=3

I hope this helps.
P.S. I am trying to start up my own homework help website. I would be extremely grateful if you would e-mail me some feedback on the help you received to adam.chapman@student.manchester.ac.uk