If #2x^2 – 3xy + y^2 + x + 2y – 7 = 0#, then #dy/dx# is equal to –?

Ans: #(3y-4x-1)/(2y-3x+2)#

1 Answer
Tom B.
Jan 2, 2018

the answer as given is : #(3y-4x-1)/(2y-3x+2)#

Explanation:

derive the equation first:
#4x*dx -(3y*dx+3x*dy)+2y*dy+1+2dy = 0#
rewrite it:
#dy(-3x+2y+2) +4x*dx-dy*dx+1=0#
rewrite:
#dy(2y-3x+2)= dx(3y-4x-1)#
divide by dx:
#dy/dx *(2y-3x+2)= (3y-4x-1)#
divide by 2y-3x+2
#dy/dx= (3y-4x-1)/(2y-3x+2)#
and you have the final answer

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