Question

Solve the system of equations
y(x+y)^2=9
y(x^3-y^3)=7

Answer 1

need help??

Answer 2

It's just a challenge, I don't need help. =P

Answer 3

lol :P

Answer 4

aeeooo ;P
let's see lol

Answer 5

I'll make it easier by saying I'm only looking for integer solutions.

Answer 6

@Kainui you have alot of math medals why not claim your titles

Answer 7

trying \(x,y=1,2,3\) by trial and error

Answer 8

I'm really not interested in seeing people get the answer, I have the answer. I just like to see how people try to solve problems since it helps me learn new ways to think. Whether they work or not isn't important since they might not work to solve this but maybe they work to solve other things. =)

Answer 9

\(\large \color{black}{\begin{align} y(x+y)^2&=9 \hspace{.33em}\\~\\
y(x^3-y^3)&=7 \hspace{.33em}\\~\\
\dfrac{y(x+y)^2}{y(x^3-y^3)}&=\dfrac{9}{7}\hspace{.33em}\\~\\
\dfrac{(x+y)^2}{(x^3-y^3)}&=\dfrac{9}{7}\hspace{.33em}\\~\\
\dfrac{(x+y)^2}{(x-y)(x^2+xy+y^2)}&=\dfrac{9}{7}\hspace{.33em}\\~\\
\dfrac{(x+y)^2}{(x-y)(x^2+2xy+y^2-xy)}&=\dfrac{9}{7}\hspace{.33em}\\~\\
\dfrac{(x+y)^2}{(x-y)((x+y)^2-xy)}&=\dfrac{9}{7}\hspace{.33em}\\~\\
\dfrac{(x-y)((x+y)^2-xy)}{(x+y)^2}&=\dfrac{7}{9}\hspace{.33em}\\~\\
(x-y)\left(1-\dfrac{xy}{(x+y)^2}\right)&=\dfrac{7}{9}\hspace{.33em}\\~\\
(x-y)-\dfrac{xy(x-y)}{(x+y)^2}&=1-\dfrac{2}{9}\hspace{.33em}\\~\\
\end{align}}\)

now if u assume \(x-y=1\) for integers and thus \(\dfrac{xy}{(x+y)^2}=\dfrac{2}{9}\) u can get a solution

Answer 10

.

Answer 11

Ahh interesting, yeah I like how you're juggling around the binomials.

Answer 12

I actually really like that last step now that I'm reading it through, since you reasoned out that x and y are consecutive numbers and that multiplying them together is 2 and their sum squared is 9 is pretty obvious what the answer is, that's super nice actually haha.

Answer 13

Also note that we have only \(4\) cases here
\(\large \color{black}{\begin{align}
(x-y)\left(1-\dfrac{xy}{(x+y)^2}\right)&=1\times \dfrac{7}{9}\hspace{.33em}\\~\\
(x-y)\left(1-\dfrac{xy}{(x+y)^2}\right)&=\dfrac{1}{3}\times \dfrac{7}{3}\hspace{.33em}\\~\\
(x-y)\left(1-\dfrac{xy}{(x+y)^2}\right)&=\dfrac{7}{3}\times \dfrac{1}{3} \hspace{.33em}\\~\\
(x-y)\left(1-\dfrac{xy}{(x+y)^2}\right)&=\dfrac{7}{9}\times 1 \hspace{.33em}\\~\\
\end{align}}\)

Answer 14

\(\rm {(x-y)(x^2+2xy+y^2-xy)}\)


-xy ???

Answer 15

i mean what is ur question lol

Answer 16

My way of doing this was kinda brute-forcey and based around this equation:

y(x+y)^2=9

This immediately implies either x+y=+-3 and y=1 or y=9 and x+y=+-1. From there I just guessed around.

Answer 17

that gives this

\(\large \color{black}{\begin{align}
y(x+y)^2=9 \hspace{.33em}\\~\\
\implies (x+y)=\pm \dfrac{3}{\sqrt y} \hspace{.33em}\\~\\
\end{align}}\)

Answer 18

Yeah, since we know x and y are integers we essentially know y is positive and a square, which limits it to 1 or 9.

Answer 19

srry for late reply >lag< >.<
okay so how did you get -xy ??

Answer 20

we initially had \((x-y)(x^2+xy+y^2) \)


\(\large \color{black}{\begin{align}
(x-y)(x^2+xy+y^2) \hspace{.33em}\\~\\
\implies (x-y)(x^2+\color{red}{2}xy-\color{red}{1}+y^2) \hspace{.33em}\\~\\
\end{align}}\)

it is the same