Level to a square

Here, it is given that

x + y = 9

x^2 + y^2 = 41

Now , The formulae for x^2 + y^2 are:

(( x + y ))^2 - 2xy

(( x - y ))^2 + 2xy

Here, we are given the value for x + y NOT x - y.

So, we have to use the first formula.

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

Substitute the values

(9)^2 - 2xy = 41

81 - 2xy = 41

81 = 41 + 2xy

2xy = 81 - 41

2xy = 40

xy = 40 / 2

xy = 20

Hence we get 20 as our answer

$x+y=9\cdots (1)$

$x^2+y^2=41\cdots (2)$

From $(2)$ ,

$x^2+y^2+2xy-2xy=41$

$(x+y)^2-2xy=41\cdots (3)$

Sub $(1)$ into $(3)$ :

$9^2-2xy=41$

$81-2xy=41$

$2xy=40$

$xy=\boxed{20}.$

the simplest way to solve this is find the possibilities of 2 numbers adding up to 9. there are 4 possibilities, they are:

1+8=9

2+7=9

3+6= 9

4+5=9.

Now, add the squares of the above numbers,

1+64=65

4+49=53

9+36=45

16+25=41

Only 4 and 5 satisfy the conditions,

x+y=9 and

x^2+y^2=41

thus substitute the values in, xy

"""""""""""""""xy=4 x 5=20""""""""""""""""""""""""

Easier solution than Anubhav Sharma.

First, square both sides of the first equation to obtain (x+y)^2=9^2=81. (x+y)^2 can also be written as x^2+2xy+y^2 after multiplying using FOIL or the binomial theorem.

So we have x^2+2xy+y^2=81 and x^2+y^2=41.

Subtracting the second equation from the first, we obtain x^2+2xy+y^2-(x^2+y^2)=81-41=40

x^2+2xy+y^2-(x^2+y^2)=2xy+(x^2+y^2)-(x^2+y^2)=2xy=40

Finally, since 2xy=40, dividing by 2 on both sides we obtain xy=20 .

(x+y)^2 =(9)^2 x^2+y^2+2xy =81 x^2+y^2=41 41+2xy=81 2xy= 40 xy=20

xy=((x+y)^2-(x^2+y^2))/2

$\begin{aligned} {(x + y)}^2 & = x^2 + y^2 + 2xy\\ 9^2 = 41 + 2xy\\ 2xy = 81-41\\ xy = \dfrac{40}{2}\\ xy = \color{#69047E}{\boxed{20}} \end{aligned}$

We have: $x+y=9$ ⇒ $(x+y)^{2}= 9^{2}=81$

In the other hand, $(x+y)^{2}= x^{2}+2xy+y^{2}$

But $x^{2}+y^{2}=41$ , and we also can replace $81$ to $(x+y)^{2}$

⇒ $81=41+2xy$

$2xy= 40$

$xy= \boxed{20}$

(x+y)^2-2xy =41 xy=(81-41)/2 xy=20

I ended up squaring the left and right sides of the first equation to get

x^2+2xy+y^2 = 81 x^2+y^2=81-2 x y x^2+y^2 = 41 = 81-2 x y 2xy = 40 xy=20

Given,

---

$x+y = 9$

---

$x^{2}+y^{2} = 41$

---

We know that,

$[x+y]^{2} = x^{2}+y^{2} +2xy$

---

Substituting the values we get,

---

$9^{2} = 41 + 2xy$

---

$81 - 41 = 2xy$

---

$40 = 2xy$

---

$\frac{40}{2} = xy$

---

$20 = xy$ = $Answer$

square both the sides .......then substitute the values given......

Brute force, using Python sympy library:

>>> from sympy import *

>>> x,y=symbols('x y')

>>> solve(Eq(x+y-9),y)

[-x + 9]

>>> (x 2+y 2-41).subs(y,_[0])

x 2 + (-x + 9) 2 - 41

>>> _.expand()

2 x 2 - 18 x + 40

>>> solve(Eq(_),x)

[4, 5]

>>> for val in _:

... (-x+9).subs(x,val)

...

5

4

I did trial and error. All possibilities for X and Y (Not decimals): 1^2 + 8^2 = 65 2^2 + 7^2 =51 3^2 + 6^2 =45 4^2 + 5^2 =41 * * After concluding that 5 and 4 represent X and Y, I multiplied them together to get 20.

Squaring both sides we get (X+Y)^ 2=9^2 that is X^2 +Y^2 + 2XY =81, Substituti9ng values of X+Y we get 41 + 2XY = 81 ,so XY =20 Ans

K.K.GARG,India

using (a+b)^2 we can get hte value of xy..