How to find the amount of gold of a bracelet when it comes from two different alloys?

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The problem is as follows:

Vincent orders a gold bracelet to Tiffany's to be casted on the occasion of his wedding anniversary to Jenny. In order to comply with the order, the master smith melts a gold bar which has a millesimal fineness equal to $0.8$ with another two gold bars of the same weight as the initial bar but with different fineness between them. Once the bracelet is presented to Vincent to get his approval, he decides the end product is not to his liking, so its returned to the master smith with a note ordering him to add $20\,g$ of pure gold to the bracelet. A later inspection at the quality control lab revealed that this procedure made the bracelet to have only $14\%$ of impurities. If the other two ingots had been melted by separate, $\textrm{18 karat}$ gold would have been obtained. Find the weight in grams of the bracelet which was commisioned by Vincent.

The alternatives given in my book are as follows

$\begin{array}{ll} 1.&45\,g\\ 2.&48\,g\\ 3.&50\,g\\ 4.&54\,g\\ \end{array}$

I'm confused exactly how to solve this problem. What I assumed is that the weight of the bracelet could be found by using an expression relating the concentration of gold by the ingolts involved. Assuming that in the equation I'm adding the amount of gold. But is that rationale correct?.

Let $x$: weight of the ingot

$x(f_1)+x(f_2)+x(0.8)=(3x)(f_3)$

After adding $20$ grams of pure gold this results into:

$3x(f_3)+20(1)=(3x+20)(0.86)$

The third part is some confusing regarding melting the other two ingots by separate, what's intended?. Does it mean melting those ingots without the bracelet and adding the pure gold?.

I assumed this as:

$x(f_1)+x(f_2)=2x\left(\frac{18}{24}\right)$

In this given situation the weight of the bracelet is given by:

$3x+20$

Thus the only that it is left is to find who's x: But there aren't sufficient equations to find that value, can someone help me here?.

The equations would look like:

$x(f_1)+x(f_2)+x(0.8)=(3x)(f_3)$

Then

$f_1+f_2+0.8=3f_3$

$f_1+f_2=2\left(\frac{3}{4}\right)=\frac{3}{2}$

$3x(f_3)+20(1)=(3x+20)(0.86)$

But as mentioned this results in a system which cannot be solved, can someone help me?. What is the way to solve this sort of problem involving mixtures?.