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I couldn't solve the above equation. Below, I describe my attempt at solving it.

$$x^2 + y^2 = 10 \tag{1}$$ $$\frac{1}{x} + \frac{1}{y} = \frac{4}{3} \tag{2}$$ Make the denominator common in the RHS of $(2)$. $$\frac{y + x}{xy} = \frac{4}{3} \tag{2.1}$$ Multiply $(2.1)$ through by $3$: $$\frac{3(y + x)}{xy} = 4 \tag{2.2}$$

Let $y = mx$
Substitute $y = mx$ into $(1)$ and $(2.2)$: $$x^2 + m^2x^2 = 10 \tag{1.1}$$ $$\frac{3(mx + x)}{mx^2} = 4$$ Factorise: $$\frac{3x(m + 1)}{x(mx)} = 4$$ $$\frac{3(m + 1)}{(mx)} = 4$$ Cross multiply: $$3(m+1) = 4mx$$ Collect like terms: $$4mx - 3m = 3$$ Factorise: $$m(4x - 3) = 3$$ Divide through by $(4x - 3)$: $$m = \frac{3}{4x - 3} \tag{3}$$

Substitute $(3)$ into $(1.1)$ $$x^2 + \left(\frac{3}{4x - 3}\right)^2x^2 = 10$$ $$x^2 + \frac{9x^2}{16x^2 - 24x + 9} = 10$$ Multiply through by ${16x^2 - 24x + 9}$: $$16x^2 - 24x^3 + 9x^2 + 9x^2 = 10(16x^2 - 24x + 9)$$ Divide through by $2$: $$8x^4 - 12x^3 + 9x^2 = 80x^2 - 120x + 45$$ $$8x^4 - 12x^3 - 71x^2 +120x - 45 = 0$$. $x = 0$ is not a solution of the above equation, so divide through by $x^2$: $$8x^2 - 12x - 71 + \frac{120}{x} - \frac{45}{x^2} = 0 \tag{1.2}$$ Let $v = x - \frac{k}{x}$.
$$v^2 = x^2 - 2k + \frac{k^2}{x^2}$$

Rewriting $(1.2)$: $$\left(8x^2 - 71 - \frac{45}{x^2}\right) + \left(-12x + \frac{120}{x}\right)$$ $$\left(8x^2 - 71 - \frac{45}{x^2}\right) + \left(-12(x - \frac{10}{x}\right) \tag{1.3}$$

Putting $k = 10$, works for $v$, but not for $v^2$. I don't know any other approach to solving the equation from $(1.3)$.

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  • $\begingroup$ If $x, y \in \mathbb{Z}$ then by inspection $(\pm x, \pm y) = (\pm1, \pm 3)$ $\endgroup$
    – user284001
    Commented Jul 28, 2017 at 13:31
  • $\begingroup$ In the beginning, notice that $x+y=\frac 43xy$ and $x^2+y^2=(x+y)^2-2xy$. This will allow you to find both $xy$ and $x+y$ explicitly. And once you know them, you know $x$ and $y$ via the usual method. $\endgroup$
    – user228113
    Commented Jul 28, 2017 at 13:32
  • $\begingroup$ @G.Sassatelli thanks, I'll add the answer myself once I have the time. $\endgroup$ Commented Jul 28, 2017 at 13:34

4 Answers 4

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The right approach is via symmetric functions: set $s=x+y$, $p=xy$. Then

  • $x^2+y^2=s^2-2p,\:$ so we have the relation: $\;s^2-2p=10 \tag{1}$
  • $\dfrac 1x+\dfrac 1y=\dfrac sp=\dfrac43,\:$ whence a second relation: $\;3s=4p\tag{2}$.

Relation $(1)$, taking relation $(2)$ into account, yields a quadratic equation in $s$: $$2s^2-3s-10.$$ Solve for $s$, then $p$, and it comes down to the high school classical problem of finding two numbers, given their sum and their product.

You should find $4$ pairs of solutions corresponding to the fact that two conics intersect in $4$ points.

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  • $\begingroup$ Thank you very much for the pointer. $\endgroup$ Commented Jul 29, 2017 at 9:08
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from the second equation we get $$x+y=\frac{4}{3}xy$$ and $$x,y\ne 0$$ solving this equation for $y$ we get $$y=\frac{x}{\frac{4}{3}x-1}$$ this can we insert in the first equation $$x^2+\left(\frac{x}{\frac{4}{3}x-1}\right)^2=10$$ simplifying and factorizing we get $$2\, \left( x-1 \right) \left( x-3 \right) \left( 8\,{x}^{2}+20\,x-15 \right) =0$$ can you finish?

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  • $\begingroup$ Yes I can. thank you. :) $\endgroup$ Commented Jul 28, 2017 at 13:43
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Let $s=x+y$ and let $p=xy$. Then you know that $10=x^2+y^2=(x+y)^2-2xy=s^2-2p$ and that $\frac43=\frac1x+\frac1y=\frac sp$. So let us solve the system$$\left\{\begin{array}{l}s^2-2p=10\\p=\frac34s.\end{array}\right.$$Replacing $p$ by $\frac34s$ in the first equation gives $s^2-\frac32s=10$. This equation has two solutions: $s=4$ (for which $p=3$) and $s=-\frac52$ (for which $p=-\frac{15}{8}$). Now, all that's left is to solve the systems$$\left\{\begin{array}{l}x+y=4\\xy=3\end{array}\right.\text{ and }\left\{\begin{array}{l}x+y=-\frac52\\xy=-\frac{15}8.\end{array}\right.$$

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help from the graph : $$x^2+y^2=10 \to \text {Circle}$$

$$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{3}\to y=\dfrac{3x}{4x-3}\to \text{Hemographic }$$

enter image description here

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  • $\begingroup$ Did you mean hyperbola? $\endgroup$ Commented Jul 28, 2017 at 14:49
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    $\begingroup$ does this help to solve the problem? $\endgroup$
    – MAN-MADE
    Commented Jul 28, 2017 at 15:04

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