Question

The answer to the problem

Answer 1

Hi,

Solving:


`\frac{2 {y}^(2) - 6y - 20}{4y + 12} / \frac{ {y}^(2) + 5y + 6}{ 3 {y}^(2) + 18y + 27 } \\ \frac{2 {y}^(2) - 6y - 20}{4y + 12} / \frac{3 {y}^(2) + 18y + 27 }{{y}^(2) + 5y + 6} \\ \frac{2( {y}^(2) - 3y - 10)}{4y + 12} / \frac{3 {y}^(2) + 18y + 27}{ {y}^(2) + 5y + 6} \\ \frac{2( {y}^(2) - 3y - 10}{2(2y + 6)} * \frac{3 {y}^(2) + 18y + 27}{ {y}^(2) + 5y + 6} \\ \frac{ \\ot2( {y}^(2) - 3y - 10) }{ \\ot2(2y + 6)} * \frac{3( {y}^(2) + 6y + 9) }{ {y}^(2) + 5y + 6} \\ \frac{ {y}^(2) + 2y - 5y - 10 }{2(y + 3)} * \frac{3( {y}^(2) + 6y + 9) }{ {y}^(2) + 3y + 2y + 6} \\ (y * (y + 2) - 5(y + 2))/(2(y + 3)) * (3(y + 3)^(2) )/(y * (y + 3) + 2(y + 3)) \\ ( (y + 2) * (y - 5))/(2(y + 3)) * (3(y + 3)^(2) )/((y + 3) * (y + 2)) \\ (y - 5)/(2(y + 3)) * 3(y + 3) \\ (y - 5)/(2) * 3 = (3y - 15)/(2) \: \: \: \: \: \: \: \: \: \: \: \: result`

Answer: B

Hope this helps.
r3t40