Question 992486
We divide 16 into two parts/numbers: {{{x}}} and {{{y}}} , so that
{{{x+y=16}}} , and we say that
{{{x}}}= the bigger number, and
{{{y}}}= the smaller number.
It is not explicitly stated, but I suspect that the wording implies that {{{x}}} and {{{y}}} are both positive numbers, and probably positive integers.
The phrase "the square of the bigger number is 64 more than the square of the smaller number" translates as
{{{x^2=y^2+64}}}<--->{{{x^2-y^2=64}}} .
Now we have a system of two equations:
{{{system(x^2-y^2=64,x+y=16)}}} or {{{system(x^2=y^2+64,x+y=16)}}} .
From there, we can take different roads to the solution.


ONE WAY:
{{{system(x^2-y^2=64,x+y=16)}}}--->{{{system((x+y)(x-y)=64,x+y=16)}}}--->{{{system((x+y)(x-y)/(x+y)=64/16,x+y=16)}}}--->{{{system(x-y=4,x+y=16)}}}--->{{{system(x-y+x+y=4+16,x+y=16)}}}--->{{{system(2x=20,x+y=16)}}}--->{{{system(x=10,x+y=16)}}}--->{{{system(x=10,10+y=16)}}}--->{{{system(x=10,y=16-10)}}}--->{{{highlight(system(x=10,y=6))}}}


ANOTHER WAY:
{{{system(x^2=y^2+64,x+y=16)}}}--->{{{system(x^2=(16-x)^2+64,y=16-x)}}}--->{{{system(x^2=256-32x+x^2+64,y=16-x)}}}--->{{{system(0=320-32x,x+y=16)}}}--->{{{system(32x=320,x+y=16)}}}--->{{{system(x=320/32,x+y=16)}}}--->{{{system(x=10,x+y=16)}}}--->{{{system(x=10,10+y=16)}}}--->{{{system(x=10,y=16-10)}}}--->{{{highlight(system(x=10,y=6))}}}


A "CHEATING" WAY:
{{{x^2=y^2+64}}}<--->{{{x^2=y^2+8^2}}}
A right triangle with legs measuring {{{y}}} and {{{8}}} has a hypotenuse measuring {{{x}}} .
Look up a table of Pythagorean triples.
In that table you would find, in the first two columns,
{{{8}}} and {{{y}}} (not necessarily in that order.
The third column will show {{{x}}} .
In the first few rows you find two choices:
{{{matrix(2,3,6,8,10,8,15,17)}}}
The first choice gives
{{{y=6,x=10,x+10=6+10=16)}}} and shows you the solution.
The other row shows
{{{y=15,x=17,x+y=15+17=32)}}} and has nothing to do with the solution.