Need to simplify. I keep getting 1, but apparently that is wrong. Thanks in advance!
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Need to simplify. I keep getting 1, but apparently that is wrong. Thanks in advance!
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$\displaystyle \frac{9{y^2}-12y+4}{(3y-2)^2} = \frac{9{y^2}-12y+4}{9{y^2}-12y+4} = 1 $
OR
$\displaystyle \frac{9{y^2}-12y+4}{(3y-2)^2} = \frac{(3y-2)^2}{(3y-2)^2} = 1$
$\displaystyle (a-b)^2 = {a^2}-2ab+{b^2}$
How can this be wrong?! 1 IS the asnwer
To check your answer, try plugging $\displaystyle y=1$ in your fraction, and you'll see that the result is satisfied!
Well, if you want to be really hard nosed about it, $\displaystyle \frac{y^2- 12y+ 4}{(3y-2)}$ is equal to 1 for all y except y= 2/3 where it is undefined. The functions $\displaystyle f(y)\frac{y^2- 12y+ 4}{(3y-2)}$ and g(y)= 1 are NOT the same- they have different natural domains.