Any help with these would be greatly appreciated thank you! 6.) (x^2 - 36) * [(3x + 1) / (x + 6)] (y^2 - 9) / [(y + 3) / (y - 3)]
Follow Math Help Forum on Facebook and Google+
Originally Posted by iDrum Any help with these would be greatly appreciated thank you! 6.) (x^2 - 36) * [(3x + 1) / (x + 6)] (y^2 - 9) / [(y + 3) / (y - 3)] You need to use the difference of 2 squares rule on both problems. It is $\displaystyle a^2-b^2= (a-b)(a+b)$ Here's a kick start $\displaystyle x^2 - 36 = x^2 - 6^2 = (x-6)(x+6) $
Originally Posted by iDrum Any help with these would be greatly appreciated thank you! 6.) (x^2 - 36) * [(3x + 1) / (x + 6)] (y^2 - 9) / [(y + 3) / (y - 3)] rewrite $\displaystyle \frac {y^2-9}{1}\ \frac{y-3}{y+3}$ then $\displaystyle \frac {(y+3)(y-3)}{1}\ \frac{y-3}{y+3}$ see the rest?
View Tag Cloud