Question 259570
Let X and Y be the numbers. We get
(i) {{{x-y = 8}}}
(ii) {{{x^2+y^2 = 320}}}
solve (i) for x to get
(iii) {{{x = 8 + y}}}
substitute (iii) into (ii) to get
(iv) {{{(8+y)^2 + y^2 = 320}}}
expanding, we get
(v) {{{y^2 + 16y + 64 + y^2 = 320}}}
combine like terms to get
(vi) {{{2y^2 + 16y + 64 = 320}}}
setting =0 and factoring, we get
(vii) {{{2y^2 + 16y -256 = 0}}}
factored, we get
(viii) {{{(y+16)(y-8)=0}}}
solving for y, we get
y = -16  or y = 8.
So, we have (-8, -16) ; (16,8)