<s>[INST] $x^2 + y^2 = 10 , \,\, \frac{1}{x} + \frac{1}{y} = \frac{4}{3}$ 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 $(1.3)$.
 [/INST] help from the graph :
$$x^2+y^2=10 \to \text {Circle}$$
$$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{3}\to y=\dfrac{3x}{4x-3}\to \text{Hemographic
}$$

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