# need a hintx+y=5 xy=3 x^2+y^2=? what means 'conventional way' to solve?

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x+y=5

xy=3

x^2+y^2=?

Remember that (x+y)^2= x^2+2xy+y^2.

Now let's substitute the following values.

(5)^2=x^2+2(3)+y^2

25=6+x^2+y^2

Let's subtract 6:

19=x^2+y^2

x+y=5

xy=3

x^2+y^2=?

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**

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**