need a hintx+y=5 xy=3 x^2+y^2=? what means 'conventional way' to solve?
That means that you will need to us the first two equations to solve for the third.
Let us rewrite the third equation.
==> We know that ( x+y)^2 = x^2 + 2xy + y^2
==> x^2 + y^2 = (x+y)^2 - 2xy
Now we will substitute the values in the first two equation into the equivalent to the third equation.
==> x^2 + y^2 = (5)^2 - 2(3) = 25 - 6 = 19
==> x^2 + y^2 = 19
Remember that (x+y)^2= x^2+2xy+y^2.
Now let's substitute the following values.
Let's subtract 6:
Solving "conventional way" means to determine x or y from the first equation and to substitute it into the second.
But, the easier way to solve the problem would be to raise to square both sides, the sum x + y = 5:
(x + y)^2 = 5^2
We'll expand the square:
x^2 + xy + y^2 = 25
We'll keep the sum of the squares to the left side:
x^2 + y^2 = 25 - xy
But xy = 3
x^2 + y^2 = 25 - 3
x^2 + y^2 = 22