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Solve. \[ \begin{array}{l}\frac{3}{y+5}+\frac{7}{y^{2}-5 y+25}=\frac{225}{y^{3}+125} \\ \text { Select the correct choice below and, if necessary, fill in the answer box to complete your choice. } \\ \text { A. The solution set is }\{y \mid y \text { is a real number and } y \neq \square\} \text {. } \\ \text { (Type an integer or a simplified fraction. Use a comma to separate answers as needed.) } \\ \text { B. The solution(s) is/are } \square \text {. } \\ \text { (Type an integer or a simplified fraction. Use a comma to separate answers as needed.) } \\ \text { C. There is no solution. }\end{array} \text { ? } \]

Ask by Macdonald Brooks. in the United States
Feb 26,2025

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The solution is \( y = \frac{23}{3} \).

Solución

We start with the equation   3/(y + 5) + 7/(y² − 5y + 25) = 225/(y³ + 125). Notice that the denominator on the right factors as a sum of cubes:   y³ + 125 = (y + 5)(y² − 5y + 25). Thus, the equation becomes   3/(y + 5) + 7/(y² − 5y + 25) = 225/[(y + 5)(y² − 5y + 25)]. Multiply both sides by (y + 5)(y² − 5y + 25) to eliminate the denominators (noting that y ≠ -5 and y² − 5y + 25 ≠ 0):   3(y² − 5y + 25) + 7(y + 5) = 225. Expand and simplify:   3y² − 15y + 75 + 7y + 35 = 225,   3y² − 8y + 110 = 225. Subtract 225 from both sides:   3y² − 8y − 115 = 0. Next, solve the quadratic equation using the quadratic formula:   y = [8 ± √(64 + 1380)]/(2*3),   y = [8 ± √1444]/6,   y = [8 ± 38]/6. This gives two potential solutions:   y = (8 + 38)/6 = 46/6 = 23/3,   y = (8 − 38)/6 = −30/6 = −5. However, y = −5 must be rejected since it makes the denominator (y + 5) equal to zero in the original equation. Thus, the only valid solution is y = 23/3. The correct choice is:   B. The solution(s) is/are 23/3.

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Bonus Knowledge

To solve the equation \[ \frac{3}{y+5}+\frac{7}{y^{2}-5 y+25}=\frac{225}{y^{3}+125}, \] we first factor the denominators. The quadratic \(y^2 - 5y + 25\) does not factor nicely, so we will complete the square or use it as is. The cubic \(y^3 + 125\) can be factored as \((y + 5)(y^2 - 5y + 25)\). Now the equation becomes: \[ \frac{3}{y+5} + \frac{7}{(y - \frac{5}{2})^2 + \frac{75}{4}} = \frac{225}{(y + 5)(y^2 - 5y + 25)}. \] Multiplying both sides by \((y + 5)(y^2 - 5y + 25)\) to eliminate the denominators: \[ 3(y^2 - 5y + 25) + 7(y + 5) = 225. \] Expanding and simplifying the left-hand side: \[ 3y^2 - 15y + 75 + 7y + 35 = 225. \] \[ 3y^2 - 8y + 110 = 225. \] Subtract 225 from both sides: \[ 3y^2 - 8y - 115 = 0. \] Now we can solve this quadratic equation using the quadratic formula: \[ y = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-115)}}{2 \cdot 3} = \frac{8 \pm \sqrt{64 + 1380}}{6}. \] \[ y = \frac{8 \pm \sqrt{1444}}{6} = \frac{8 \pm 38}{6}. \] This yields two solutions: \[ y_1 = \frac{46}{6} = \frac{23}{3}, \quad y_2 = \frac{-30}{6} = -5. \] However, we need to check for extraneous solutions by considering the restrictions imposed by the original denominators. The term \(y + 5\) in the denominator indicates that \(y\) cannot equal \(-5\). Hence, this value must be excluded from the solution set. Thus, the solution set is **A. The solution set is** \(\{y \mid y \text{ is a real number and } y \neq -5\}.\)

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