XY Problem

See $30={ x }^{ 2 }+{ y }^{ 2 }=(x+y{ ) }^{ 2 }-2xy$ , as $x+y=10$ ; we know $(x+y{ ) }^{ 2 }-2xy=100-2xy=30$ . Thus $xy=\boxed { 35 }$ .

$\begin{aligned} (x+y)^2&=x^2+y^2+2xy\\{10}^2&=30+2xy\\70&=2xy\\35&=xy \end{aligned}$

Has anyone noticed that this system of equations has no solution for real numbers $x,y$ ? Seems like kind of a flaw.

Completing the square:

x^2 + 2xy + y^2 = 30 + 2xy

(x + y)^2 = 30 + 2xy

x + y = 10

100 = 30 + 2xy

2xy = 70

xy = 35

(x+y) ^2= 10^2
x^2+ y^2 + 2xy = 100,
30 + 2xy = 100,
2xy = 100-30
2xy = 70
thus, xy = 35

given the following equation ,
x+y=10, x^+y^=30, ( ^=means power of 2) the formula is (x+y)^=x^+y^+2xy,
10x10=30+2xy 100 = 30+2xy 70=2xy 35=xy

the answer is 35.

(x + y)^2= x^2 + y^2 + 2 .x. y

or,(10)^2 = 30 + 2xy

or, 70 =2xy

or, 35 =xy

$X^{2}$ + $Y^{2}$ = $(X+Y)^{2}$ -2xy

By substituting the given values, we get

$10^{2}$ -2xy =30

$\Rightarrow$ 100-2xy=30

$\Rightarrow$ 2xy=100-30

$\Rightarrow$ 2xy=70

$\Rightarrow$ xy=35

**therefore the value of xy= $\boxed{35}$

If (x + y)^2 = x^2 + 2xy + y^2 Then (10)^2 = (x^2 + y^2) + 2xy; (10)^2 = (30) + 2xy; 2xy = 70; xy = 35

The answer to this question would be $2xy = (x+ y)^2 - (x^2 + y^2) = 70 \ \ \therefore\ \ xy = 35.$

Note 1: It is interesting to take this as starting point for finding the values of $x$ and $y$ themselves. Here is how I proceed: $(x - y)^2 = (x^2 + y^2) - 2xy = (x^2 + y^2) - \left[(x+y)^2 - (x^2 + y^2)\right] \\ = 2(x^2 + y^2) - (x+y)^2 = 2\cdot 30 - 10^2 = -40$ Now we know that $\begin{cases} x + y = 10 \\ x - y = \pm\sqrt{-40} = \pm2\sqrt{10}i \end{cases}$ We find $x$ and $y$ by adding or subtracting these equations, then halving: $x,y = \frac{10 \pm 2\sqrt{10}i}2 = 5 \pm \sqrt{10}i.$

Note 2: Others have said this too: there are no real solutions to this equation. We see this very quickly by realizing that for real numbers, given the sum $x+y$ , the value $x^2 + y^2$ is minimal when $x = y$ . In this case, this implies that $x^2 + y^2 \geq 2\cdot 5^2 = 50$ . Because that is not true here, there are no real solutions.

x^2+y^2= 30<br> x+y=10<br> <br> y=10-x<br> <br> x^2+(10-x)^2= 30<br> x^2+100-20x+x^2= 30<br> 2x^2-20x+100=30<br> 2x^2-20x+70<br> x^2-10x+35=0<br>

We now that the standard root form of a quadratic equation is always

ax^2-b(sum)*x+(product)=0

Therefore, xy=35

x+y= 10 square both side (x+y)^2=100 x^2 + y^2 + 2xy =100 30+2xy = 100 xy= 100-30 / 2 xy = 70/ 2 xy = 35

(x^2 )+(y^2)=30; x+y=10; (x+y)^2=100; 100=(x^2)+(2xy)+(y^2); 100=30+2xy; 70=2xy; 35=xy . No big deal. You can find out the answer using the formula: (a+b)^2=(a^2)+2ab+(b^2)

We know $x^{2} + 2xy + y^{2} = (x + y)^{2} = 10^{2} = 100$ . Hence, $2xy = 100- (x^{2} + y^{2}) = 100 - 30 = 70$ . Therefore, $xy = \frac{70}{2} = \boxed{35}$ .

(x+y)^2=(x^2)+(y^2)+2xy=30+2xy=10^2=100 Thus xy=35. As simple as that!

x^2+y^2=30 x+y=10,squaring on both sides we get, x^2+y^2+2xy=100; so 30+2*xy=100; 2xy=70; xy=35.

taking x+y=10
(x+y)^2=10^2
x^2+y^2+2xy=100
30+2xy=100 [x^2+y^2=30]
2xy=70
xy=35

(x+y) ²=x²+y²+2xy 10²=30+2xy 2xy=100-30 2xy=70 xy=35

x+y=10

x^2 + 2(x)(y) + y^2 = 100

30+ 2 (x) (y) = 100

xy= (100-30)/2 = 35

x²+y²=30; x+y=100; xy=?

(x+y)²=10² => x²+2xy+y²=100;

x²+y²+70=30+70 => x²+y²+70=100;

x²+2xy+y²=x²+y²+70 => 2xy=x²+y²+70-x²-y² => 2xy=70 => xy=70/2 => xy=35

We know that (x+y)² =10² --> x²+2xy+y²=100 --> xy = [100 - (x²+y²)]/2, because x²+y² = 30; --> xy = (100 - 30)/2 = 35

Squaring on both sides to x+y=10 and sub x^2+y^2=30

(x+y)^2-2xy=30 (10)^2-2xy=30 100-30=2xy →xy=35