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There are 2 bars of gold-silver alloy; one piece has 2 parts of gold to 3 parts of silver and another has 3 parts of gold to 7 parts of silver. If both bars are melted into 8 kg bar with the final gold to silver ratio of 5:11. What was the weight of the first bar?

Here's  the way I might  go about this

Call the  weight of the first bar  = A

Then the  weight of the  second  bar  =  8 - A

If  the ratio  of gold to silver in the  8 kg  bar  = 5 : 11,  then  the   weight of the  gold must  be 

(5/16) * 8  =  2.5 kg

And  the  weight  of silver  in 8kg bar   =     5.5 kg

And if  the  parts  of  gold to  silver in the  first  bar = 2 : 3,  then  gold must  be (2/5)  of the  weight of the  first  bar

And if the parts of  gold  to  silver  in the  second  bar  =  3 : 7, then gold  must  be (3/10) of the weight of the  second bar 

So  we have  this equation

(2/5) (A)  + (3/10)  (8 - A)   =    2.5      

(4/10)A  + 24/10  - (3/10)A  =  2.5

(1/10A  +  2.4  =  2.5

(1/10)A  =  1/10

A =  1  kg  =    weight of  1st  bar