Find the two numbers...?

Why on Earth do You need them expressed in radicals?

Anyway, Peter has done all essential work above:

y³ + 3y² - 13y - 38 = 0, or

y³ + 3y² + 3y + 1 - 16y - 16 - 23 = 0, or

(y+1)³ - 16(y+1) - 23 = 0

The discriminant is 23²/4 - 16³ /27 = -2101/(4*27) < 0, so according the theory there are indeed 3 real roots; let ³√ denotes cubic root, i = √(-1) as usual, then

y = ³√(23/2 + (i/2)√(2101/27)) + ³√(23/2 - (i/2)√(2101/27)) -1

x = 11 - y²

Of course You should take care according the theory to combine the 3 complex values of the 1st cubic root with the proper values of the 2nd to obtain the 3 real values for y, the rest is easy.

I sincerely hope that calendars for 2008 will be, say with beautiful girls instead of parabolas on them!

Happy New Year to everybody!

yet in any different case to look at it using substitution: enable x and y be the two numbers. we could make 2 equations from the help given: (a million) x + y = 40 5 (2) x - y = 3 positioned the two equation in terms of x or y. it somewhat is least confusing to place #2 in terms of x: (2) x = 3 + y Then substitute into the different equation: (a million) x + y = 40 5 (a million) (3 + y) + y = 40 5 (a million) 2y = 40 5 - 3 (a million) y = 21 Then returned returned into #2 (2) x = 3 + y (2) x = 3 + (21) (2) x = 24

x^2+y=7

x+y^2=11

=>

(11-y^2)^2+y-7=0

x=11-y^2

=>

y^4-22y^2+y+114=0

but you're not interested in y=3, so, dividing by y-3,

y^3+3y^2-13y-38=0

which can be solved with Cardano's method for y and thus find x (this holds the other 2 real solutions). Obviously you could have made another choice at the beginning (solved for y) and been left with a polynomial in x (still third degree though).

One note though, "trivial" would not generally imply the (2,3) solution..

lets define a as the first number and b as the second

a^2 + b = 7

b^2 + a = 11 are the 2 statements you have up there.

rearranging b^2+a=11 we have

a = 11 - b^2

a^2 + b = 7

(11 - b^2) * (11 - b^2) + b =7

121 - 22 b^2 + b^4 + b = 7

b^4 - 22 b^2 + b = -114

b = 3 in which case a = 2

Solving x^2+y=7 and x+y^2=11 simultaneously, we get x^4-14*x^2+x+38=0. This can be factored into (x-2)(x^3+2*x^2-10*x-19)=0. The solution x=2 comes from the first factor, the other three solutions satisfy x^3+2*x^2-10*x-19=0, which can be solved by a well-known formula in radicals:

x=2

or

1/6*(1268+12*I*

sqrt(6303))^(1/3)

+68/3/(1268+12*I*

sqrt(6303))^(1/3)-2/3

or

-1/12*(1268+12*I*

sqrt(6303))^(1/3)

-34/3*1/((1268+12*I*

sqrt(6303))^(1/3))-2/3

+1/2*I*sqrt(3)*(1/6*(1268

+12*I*

sqrt(6303))^(1/3)

-68/3*1/((1268+12*I*

sqrt(6303))^(1/3)))

or

-1/12*(1268+12*I*

sqrt(6303))^(1/3)

-34/3*1/((1268+12*I*

sqrt(6303))^(1/3))-2/3

-1/2*I*sqrt(3)*(1/6*(1268

+12*I*

sqrt(6303))^(1/3)

-68/3*1/((1268+12*I*

sqrt(6303))^(1/3)))

and for each x, y is 7-x^2.

First number = x

Second number = y

x^2 + y = 7

x + y^2 = 11

x = 2

y = 3

I couldn't even aply any maths at it, just too easy.

I'm assuming (2,3) is the "trivial" solution?

No