I have two equations that I am trying to simplify, please help! 1) y^2 - 6y + 9/ -(y +2 - 4) divided by y^2 - 9 / y^2 - 8y + 12 = ??? 2) 12a^2 + 26a 12 / 6 - 5a - 6a^2 = ? (I've come up with 2 (3a +2)/(3a - 2) Is this correct? ) Thanks!
I have two equations that I am trying to simplify, please help! 1) y^2 - 6y + 9/ -(y +2 - 4) divided by y^2 - 9 / y^2 - 8y + 12 = ??? 2) 12a^2 + 26a 12 / 6 - 5a - 6a^2 = ? (I've come up with 2 (3a +2)/(3a - 2) Is this correct? ) Thanks!
1) I assume that the term reads:
$\displaystyle \dfrac{\dfrac{y^2-6y+9}{-(y^2-4)}}{\dfrac{y^2-9}{y^2-8y+12}} = \dfrac{(y^2-6y+9)(y^2-8y+12)}{-(y^2-4)(y^2-9)}$
Factor numerator and denominator and cancel equal terms:
$\displaystyle \dfrac{(y^2-6y+9)(y^2-8y+12)}{-(y^2-4)(y^2-9)} = \dfrac{(y-3)(y-3)(y-2)(y-6)}{-(y+2)(y-2)(y+3)(y-3)}$
2) Same procedure as 1)
$\displaystyle \dfrac{12a^2+26a+12}{6-5a-6a^2} = \dfrac{2(2a+3)(3a+2)}{-(2a+3)(3a-2)} = \dfrac{2(3a+2)}{2-3a}$
(In your answer the negative sign in the denominator is missing)
Your result doesn't look much simpler, does it?
$\displaystyle \dfrac{(y-3)(y-3)(y-2)(y-6)}{-(y+2)(y-2)(y+3)(y-3)} = \dfrac{(y-3)(y-6)}{-(y+2)(y+3)}$
Now factor out (-1) from the first bracket of the numerator and cancel:
$\displaystyle \dfrac{(y-3)(y-6)}{-(y+2)(y+3)} = \boxed{\dfrac{(3-y)(y-6)}{(y+2)(y+3)} }$
Please: If you have a new question, start a new thread.
Your result is not correct, sorry!
1. I assume that you mean:
$\displaystyle \dfrac{2z-1}{z^2+z-6} - \dfrac{3z-5}{z^2-2z-15} + \dfrac{2z-3}{z^2-7z+10}$
1. Factor the denominator to determine the common denominator:
$\displaystyle \dfrac{2z-1}{(z-2)(z+3)} - \dfrac{3z-5}{(z+3)(z-5)} + \dfrac{2z-3}{(z-2)(z-5)}$
Therefore the common denominator is $\displaystyle d = (z-2)(z+3)(z-5)$
2. Transform the summands such that each fraction has the same denominator:
$\displaystyle \dfrac{(2z-1)\bold{\color{blue}(x-5)}}{(z-2)(z+3)\bold{\color{blue}(x-5)}} - \dfrac{\bold{\color{blue}(x-2)}(3z-5)}{\bold{\color{blue}(x-2)}(z+3)(z-5)} + \dfrac{(2z-3)\bold{\color{blue}(x+3)}}{(z-2)\bold{\color{blue}(x+3)}(z-5)}$
3. Expand the brackets in the numerator (there is now only one numerator!), collect like terms.
4. You should come out with: $\displaystyle \dfrac{z^2 + 3z - 14}{(z-2)(x+3)(z-5)}$