Get the value $x+y$

$x^4+2x^2y+y^2=(x^2+y)^2=169=(\pm 13)^2 ~~~ \Longrightarrow ~ x^2+y=\pm 13$ $x^3+y^2+x^2y+xy=(x+y)(x^2+y)=\pm 13(x+y)=91$ $\therefore ~ x+y=\pm 7$

First label the equations $x^3 + y^2 + x^2y + xy = 91 \quad (1)$ $x^4 + 2x^2y + y^2 = 169 \quad (2)$ We can factorize equation $(1)$ as $x^3 + y^2 + x^2y + xy = (x^2+y)(x+y) = 91 \quad (3)$ Similarly, equation $(2)$ gives $x^4 + 2x^2y + y^2 = (x^2 + y)^2 = 169 \Rightarrow x^2 + y = 13 \quad (4)$ Thus, substituting $(4)$ into $(3)$ gives $13*(x+y) = 91 \Rightarrow x+y = \frac{91}{13} = \fbox{7}$

$(x,y) = {-2,9} , {3,4}$ If we assume x & y positive we have very simple way $x^{3}+y^{2}+x^{2} \times y+x \times y=91$ = $(x+y) \times (x^{2}+y) = 91$ ------- (1) the second term is $(x^{2}+y)^{2} = 169$ $(x^{2}+y)= 13$ substituting in (1) we get $(x+y) = \boxed{7}$

Let $x^2=z$ then we have the second equation: $x^4+2x^2y+y^2=z^2+2zy+y^2=(z+y)^2=(x^2+y)^2=13^2$ So we get $x^2+y=13$ The first equation can be factored as follows: $x^3+y^2+x^2y+xy=x^3+xy+x^2y+y^2=x(x^2+y)+y(x^2+y)=(x^2+y)(x+y)=91$ Observe that we have a factor of $x^2+y)\ in \;the \;first\; equation \;so \;we\; can \;substitute \;its \;value \;into \;the \;equation,\;which \;becomes: 13(x+y)=91$ $x+y=\frac{91}{13}=\boxed{7}$

Given that, $x^3+y^2+x^2y+xy=91$ ....(i) and $x^4+2x^2y+y^2=169$ .....(ii)

From (ii), we get,

$x^4+2x^2y+y^2=169$

$\implies (x^2+y)^2=169 \implies x^2+y=13 \text{[Neglecting (-13) as x,y are (+ve)]}$ ....(iii)

Now, from (iii), we get $y=13-x^2$ and putting it in (i), we get ----

$x^3+(13-x^2)^2+x^2(13-x^2)+x(13-x^2)=91$

$\implies x^3+169-26x^2+x^4+13x^2-x^4+13x-x^3=91$

$\implies 13x^2-13x-78=0$

$\implies x^2-x-6=0$

$\implies (x-3)(x+2)=0 \implies x=3 \text{[Neglecting (-2) as x is (+ve)]}$

Putting $x=3$ in (iii), we get----

$y=13-3^2=13-9=4$ . Then, $x+y=3+4=\boxed{7}$

• (x^(2) + y)(x + y) = 91 (1)
• (x^(2)+ y)(x^(2) + y) = 169 (2)
• from (2)
• (x^(2)+y) = 13 (3) since x and y are positive
• sub. from (3) to (1) thin
• 13 * (x + y) = 91
• (x + y) = 91/13 = 7

x $^{3}$ +x $^{2}$ y+y $^{2}$ +xy=91

or, x $^{2}$ (x+y)+y(y+x)=91

or, (x $^{2}$ +y)(x+y)=91 - (i)

(x $^{2}$ ) $^{2}$ +2x $^{2}$ y+y $^{2}$ =169

or, (x $^{2}$ +y) $^{2}$ =169

or, x $^{2}$ +y=13 -(ii)

from (i) and (ii) 13(x+y)=91 or, x+y=7

Notice that in $x^4 + 2x^2y + y^2 = 169 = 13^2$ , the LHS (left hand side) looks a lot like the familiar $(a + b)^2 = a^2 + 2ab + b$ . Thus we find the factorization:

$x^4 + 2x^2y + y^2 = 13^2 \:\to\: (x^2 + y)^2 = 13^2 \:\to\: x^2 + y = 13$

Now we will try to factorize the first equation by grouping. Hopefully we will be able to get the expression $x^2+y$ , since we know that this equals $13$ . The first equation factorizes as so:

$x^3 + y^2 + x^2y + xy = (x + y)(x^2 + y) = (x + y)13 = 91 \:\to\: x + y = \boxed{7}$

And what a great coincidence, there is our solution!

x^3 + y^2 + x^2y + xy = 91

=>x^3 + x^2y + y^2 + xy = 91

=> x^2(x + y) + y(y + x) = 91

=>(x^2 + y)(x + y)=91.....equation 1

x^4 + 2x^2y + y^2 = 169

=> x^4 + x^2y + x^2y + y^2 = 169

=>x^2(x^2 + y) + y(x^2 + y) = 169

=> (x^2 + y)(x^2 + y) =169.....equation 1

Let x + y be A and x^2 + y be B .

so , BA= 91

B^2 = 169

B=13

Putting the value of B in equation 1

BA= 91

13A = 91

A=7 (x+y)

First, consider the first equation. The thing to note is that x cubed must be relatively small. In fact, we know that x must be less than 5, because otherwise x cubed will be 125, already larger than 91. If x is 4, that's also probably going to be too large, so we start by trying values with x as 3. Plugging in values, we see that the equation is satisfied if y equals 4. Another thing to note is that if x and y are even, then the expression is also even, so one's odd and one's even. But 3 + 4 is 7, the answer.