Question

Heeeeeeeeeeeeeeeelp solve the following Q

Answer 1

Answer:

$\huge\boxed{\sf \frac{y-5}{y-3} }$

Step-by-step explanation:

Solving it by mid-term break method:

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

Hope this helped!

~AH1807

Answer 2

Answer:

$\frac{y+5}{y-3} \textrm{ where } y \ne 8$

Step-by-step explanation:

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

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

Now we can use this to simplify the whole division:

$\frac{y^2-3y-40}{y^2-11y+24} = \frac{(y-8)(y+5)}{(y-8)(y-3)} = \frac{y+5}{y-3} \textrm{ where } y \ne 8$