質問: How do you solve #(4b+2)/(b^2-3n)+(b+2)/b=(b-1)/b#? ===================================================
回答: #n=(b*(7b+2))/9# #### Explanation: Given : #((4b+2)/(b^2-3n))+((b+2)/b)=(b-1)/b# #((4b+2)/(b^2-3n))=((b-1)/b)-((b+2)/b)# #(4b+2)/(b^2-3n)=(b-1-b-2)/b=-(3/b)# Cross multiplying, #(b)(4b+2)=(-3)(b^2-3n)# #4b^2+2b=-3b^2+9n# #9n=7b^2+2b# #n=(b*(7b+2))/9#