quadratic equation

Question 1: A bought a certainnumber of fish for 600.After 1 year,the number of fish has increased by 1.Athen decides to sell all the fish at $3 per fish less than the orginal cost.However shestill makes a profit of 27.How many fish did A buy originally?
Let x = number  of fish A boughtCost of each fish =$600/xSelling Price of each  fish = $600/x - $3Number of fish sold= x + 1Revenue = = (x +1)[ 600/x - 3]Profit = Total revenue – total cost = $27(x + 1)[ 600/x- 3] – 600 = 27(x + 1)[ 600/x - 3]= 627[ 600/x - 3] =627/(x + 1)(600/x = 627/(x +1) + 3627/(x +1) =  600/x – 3627/(x +1) =  (600 – 3x)/x 627/(x +1) =  3(200 – x)/x  209/(x +1) = (200 – x)/x  Cross multiply209x =(200 – x)(x +1)209x = (200 – x)(x+ 1)209x = -x^2 +199x +200x^2 -199x – 200 +209x = 0x^2 +10x – 200 = 0(x + 20)(x – 10) =0x = 10 A bought 10 fish originally.
2,Solve4(y/x)+4(y/x-6)=y4y      4y-- +    -----  - 24 = yx        x4y + 4y----------   = y + 4     x4y + 4y = x(y + 4)8y = xy + 4x8y –xy = 4xy (8 –x) = 4xy = 4x/(8 – x)