Question

Heeeeeeeeeeeeeeeelp solve the following Q

Answer 1

`(y+5)/(y-3) \textrm{ where } y \\e 8`

Explanation:

We'll first factorize the quadratic equations by finding their zero-points:

`y^2 - 3y - 40 = 0\\\Delta = (-3)^2 - 4\cdot 1 \cdot -40 = 9 + 160 = 169\\√(\Delta) = 13\\y = (3 \pm 13)/(2)\\y_1 = 8\\y_2 = -5\\y^2 - 3y - 40 = (y - 8)(y + 5)\\~\\y^2 - 11y + 24 = 0\\\Delta = (-11)^2 - 4\cdot 1\cdot 24 = 121 - 96 = 25\\√(\Delta)} = 5\\y = (11 \pm 5)/(2)\\y_1 = 8\\y_2 = 3\\y^2 - 11y + 24 = (y - 8)(y - 3)`

Now we can use this to simplify the whole division:

`(y^2-3y-40)/(y^2-11y+24) = ((y-8)(y+5))/((y-8)(y-3)) = (y+5)/(y-3) \textrm{ where } y \\e 8`

Answer 2

`\huge\boxed{\sf (y-5)/(y-3) }`

Explanation:

Solving it by mid-term break method:

`\displaystyle =(y^2-3y-40)/(y^2-11y+24) \\\\= (y^2 -8y+5y-40)/(y^2-3y-8y+24) \\\\= (y(y-8)+5(y-8))/(y(y-3)-8(y-3)) \\\\Take \ (y-8) \ and \ (y-3)\ common\\\\= ((y-8)(y-5))/((y-8)(y-3)) \\\\= (y-5)/(y-3) \\\\\rule[225]{225}{2}`

Hope this helped!

~AH1807