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

Question

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

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Answer 1 · 2018-01-02T17:19:07

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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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

Explanation:

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If 2x^2 – 3xy + y^2 + x + 2y – 7 = 02x^2 – 3xy + y^2 + x + 2y – 7 = 0, then dy/dxdy/dx is equal to –?

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

1 Answer

Explanation:

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Impact of this question

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

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