$x^2 + y^2 = 10 , \,\, \frac{1}{x} + \frac{1}{y} = \frac{4}{3}$ 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 $(1.3)$.
 A: 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.
A: 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?
A: 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$$\left\{\begin{array}{l}x+y=4\\xy=3\end{array}\right.\text{ and }\left\{\begin{array}{l}x+y=-\frac52\\xy=-\frac{15}8.\end{array}\right.$$
A: help from the graph :
$$x^2+y^2=10 \to \text {Circle}$$
$$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{3}\to y=\dfrac{3x}{4x-3}\to \text{Hemographic
}$$

