if `2x^2-3xy+y^2+x+2y-8=0` then `(dy)/(dx)`

Options

A. `(3y-4x-1)/(2y-3x+2)`
B. `(3y+4x+1)/(2y+3x+2)`
C. `(3y-4x+1)/(2y-3x-2)`
D. `(3y-4x+1)/(2y+3x+2)`

Answer

To find \(\frac{dy}{dx}\) for the equation \(2x^2 - 3xy + y^2 + x + 2y - 8 = 0\), we will use implicit differentiation. Here’s the step-by-step solution:

### Step 1: Differentiate both sides of the equation with respect to \(x\).

Given:
\[ 2x^2 - 3xy + y^2 + x + 2y - 8 = 0 \]

Differentiating each term:
- The derivative of \(2x^2\) is \(4x\).
- For \(-3xy\), we apply the product rule:
\[ \frac{d}{dx}(-3xy) = -3\left(x\frac{dy}{dx} + y\right) \]
- The derivative of \(y^2\) is \(2y\frac{dy}{dx}\) (using the chain rule).
- The derivative of \(x\) is \(1\).
- The derivative of \(2y\) is \(2\frac{dy}{dx}\).
- The derivative of \(-8\) is \(0\).

Putting it all together, we have:
\[ 4x - 3\left(x\frac{dy}{dx} + y\right) + 2y\frac{dy}{dx} + 1 + 2\frac{dy}{dx} = 0 \]

### Step 2: Simplify the equation.

Expanding the equation:
\[ 4x - 3x\frac{dy}{dx} - 3y + 2y\frac{dy}{dx} + 1 + 2\frac{dy}{dx} = 0 \]

Combine like terms:
\[ 4x - 3y + 1 + (-3x + 2y + 2)\frac{dy}{dx} = 0 \]

### Step 3: Isolate \(\frac{dy}{dx}\).

Rearranging gives:
\[ (-3x + 2y + 2)\frac{dy}{dx} = - (4x - 3y + 1) \]

Now, divide both sides by \((-3x + 2y + 2)\):
\[ \frac{dy}{dx} = \frac{-(4x - 3y + 1)}{-3x + 2y + 2} \]

This simplifies to:
\[ \frac{dy}{dx} = \frac{4x - 3y + 1}{3x - 2y - 2} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{4x - 3y + 1}{3x - 2y - 2} \]

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