If the sum of two numbers is 6 and the product of the two numbers is 3 then what is the sum of the reciprocals of the two numbers?

x+y=6      y = 6-x

x * y = 3

x * (6-x) = 3

-x^2+6x-3=0       x = 3 +-sqrt6        y = 3 -+sqrt6

1/(3+sqrt6)  + 1/(3-sqrt6)   =  2  

Let's take two arbitrary numbers: \(n,m\) therefore we have \(n+m = 6\) and \(nm = 3\).

So we have to find the sum of the reciprocals or: \(\frac{1}n+\frac{1}m \)

Adding them we get: \(\frac{m+n}{nm}\) using the sum and product of the two numbers we can say \(\frac{m+n}{nm} = \frac{6}3 = \boxed{2}\)

If the sum of two numbers is 6 and the product of the two numbers is 3 then
what is the sum of the reciprocals of the two numbers?

\(\text{Let number one $\mathbf{=a}$ } \\ \text{Let number two $\mathbf{=b}$ } \\ \text{Let the sum of the reciprocals of the two numbers$\mathbf{=x}$ } \)

\(\begin{array}{|rcll|} \hline x &=& \dfrac{1}{a} + \dfrac{1}{b} \quad | \quad \times ab \\\\ x\times ab &=& \left(\dfrac{1}{a} + \dfrac{1}{b} \right) \times ab \\\\ x\times ab &=& \dfrac{ab}{a} + \dfrac{ab}{b} \\\\ x\times ab &=& b + a \quad | \quad a+b = 6,\ ab=3 \\ \\ 3x &=& 6 \quad | \quad : 3 \\\\ x &=& \dfrac{6}{3} \\\\ \mathbf{x} &=& \mathbf{2} \\ \hline \end{array}\)

The sum of the reciprocals of the two numbers is 2

I have a question on this problem:\(Find \frac{x}{y} if \begin{align*} 2\sqrt{x} + \frac1y &= 13 \\ 8\sqrt{x} - \frac 3y &= 3. \end{align*}\)

If you can solve this that would be great. Thanks