Questions: Solve for (y). [ frac2y+4-frac1y+2=frac3y^2+6 y+8 ] (y=) No solution

Solution Steps

We start with the equation: \[ \frac{2}{y+4} - \frac{1}{y+2} = \frac{3}{y^2 + 6y + 8} \] First, we factor the denominator on the right-hand side: \[ y^2 + 6y + 8 = (y + 2)(y + 4) \] Thus, we rewrite the equation as: \[ \frac{2}{y+4} - \frac{1}{y+2} = \frac{3}{(y + 2)(y + 4)} \]

Next, we combine the left-hand side over a common denominator: \[ \frac{2(y + 2) - 1(y + 4)}{(y + 2)(y + 4)} = \frac{y}{(y + 2)(y + 4)} \] This simplifies to: \[ \frac{y}{(y + 2)(y + 4)} \]

Since the denominators are the same, we can equate the numerators: \[ y = 3 \]

We check if the solution \( y = 3 \) is valid by substituting it back into the original equation. The substitution confirms that both sides of the equation are equal, thus validating the solution.

Final Answer