Question 767154
You can solve for a and b uniquely and then plug those values into the expression for y OR you could be a mathematical boss and do it this way:

(a+b)^2 ... What is this really? It's just (a+b) * (a+b) = (a^2) + (2ab) + (b^2)

But hey, you already have a value for (ab) and a value for a^2+b^2!!

So if ab = 10, then 2ab = 2*10 = 20. We need not do anything for a^2+b^2, it's given to us in the form we want.

So now we just add, 20 + 30 = 50! ... Like a boss ;)

Alternative solution:

ab= 10 <=> a = 10/b
plug this value of a into second equation
(10/b)^2 + b^2 = 30
100/b^2 + b^2 = 30
(100 + b^4) = 30 * b(^2)
b^4 - 30b^2 + 100 = 0

replace b^2 with some arb variable for now, like c, so that we can use quadratic formula to solve for c, then we can solve for b

c^2 - 30c + 100 = 0 Now just use quadratic formula to find values of c
which should be c = 5 (3+sqrt(5)) and another value where c is negative, but we will ignore this one since to find b, we need to square root c, and we can't do that for negative values.

so then b = sqrt(c), from this you can solve a, and finally solve y, but this is horribly ugly and long and very very NOT boss like