# Math Help - Simplifying equations

1. ## Simplifying equations

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!

2. Originally Posted by jay1
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:

$\dfrac{\dfrac{y^2-6y+9}{-(y^2-4)}}{\dfrac{y^2-9}{y^2-8y+12}} = \dfrac{(y^2-6y+9)(y^2-8y+12)}{-(y^2-4)(y^2-9)}$

Factor numerator and denominator and cancel equal terms:

$\dfrac{(y^2-6y+9)(y^2-8y+12)}{-(y^2-4)(y^2-9)} = \dfrac{(y-3)(y-3)(y-2)(y-6)}{-(y+2)(y-2)(y+3)(y-3)}$

2) Same procedure as 1)

$\dfrac{12a^2+26a+12}{6-5a-6a^2} = \dfrac{2(2a+3)(3a+2)}{-(2a+3)(3a-2)} = \dfrac{2(3a+2)}{2-3a}$

3. Originally Posted by earboth
1) I assume that the term reads:

$\dfrac{\dfrac{y^2-6y+9}{-(y^2-4)}}{\dfrac{y^2-9}{y^2-8y+12}} = \dfrac{(y^2-6y+9)(y^2-8y+12)}{-(y^2-4)(y^2-9)}$

Factor numerator and denominator and cancel equal terms:

$\dfrac{(y^2-6y+9)(y^2-8y+12)}{-(y^2-4)(y^2-9)} = \dfrac{(y-3)(y-3)(y-2)(y-6)}{-(y+2)(y-2)(y+3)(y-3)}$

2) Same procedure as 1)

$\dfrac{12a^2+26a+12}{6-5a-6a^2} = \dfrac{2(2a+3)(3a+2)}{-(2a+3)(3a-2)} = \dfrac{2(3a+2)}{2-3a}$

Thanks! For the first equation: could the simplified answer be (y+3)(y-3)^3 / -(y+2)(y-2)^2(y-6) ?

4. Originally Posted by jay1
Thanks! For the first equation: could the simplified answer be (y+3)(y-3)^3 / -(y+2)(y-2)^2(y-6) ?
Your result doesn't look much simpler, does it?

$\dfrac{(y-3)(y-3)(y-2)(y-6)}{-(y+2)(y-2)(y+3)(y-3)} = \dfrac{(y-3)(y-6)}{-(y+2)(y+3)}$

Now factor out (-1) from the first bracket of the numerator and cancel:

$\dfrac{(y-3)(y-6)}{-(y+2)(y+3)} = \boxed{\dfrac{(3-y)(y-6)}{(y+2)(y+3)} }$

5. Okay, I got it. Thanks! How about this one: (2z - 1/z^2 + z - 6) - (3z - 5/z^2 - 2z -15) + (2z - 3/z^2 - 7z + 10) ? I've simplified to z+7/(z+3)(z-5) Is this correct?

6. Originally Posted by jay1
Okay, I got it. Thanks! How about this one: (2z - 1/z^2 + z - 6) - (3z - 5/z^2 - 2z -15) + (2z - 3/z^2 - 7z + 10) ? I've simplified to z+7/(z+3)(z-5) Is this correct?

Your result is not correct, sorry!

1. I assume that you mean:

$\dfrac{2z-1}{z^2+z-6} - \dfrac{3z-5}{z^2-2z-15} + \dfrac{2z-3}{z^2-7z+10}$

1. Factor the denominator to determine the common denominator:

$\dfrac{2z-1}{(z-2)(z+3)} - \dfrac{3z-5}{(z+3)(z-5)} + \dfrac{2z-3}{(z-2)(z-5)}$

Therefore the common denominator is $d = (z-2)(z+3)(z-5)$

2. Transform the summands such that each fraction has the same denominator:

$\dfrac{(2z-1)\bold{\color{blue}(x-5)}}{(z-2)(z+3)\bold{\color{blue}(x-5)}} - \dfrac{\bold{\color{blue}(x-2)}(3z-5)}{\bold{\color{blue}(x-2)}(z+3)(z-5)} + \dfrac{(2z-3)\bold{\color{blue}(x+3)}}{(z-2)\bold{\color{blue}(x+3)}(z-5)}$

3. Expand the brackets in the numerator (there is now only one numerator!), collect like terms.

4. You should come out with: $\dfrac{z^2 + 3z - 14}{(z-2)(x+3)(z-5)}$