Originally Posted by Ari m doesnt = -4/5 yes then what am i supposed to cancel? all the (j + 3)'s cancel. and one of the (j + 4)'s in the top cancels with the (j + 4) in the bottom
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Originally Posted by Jhevon yes all the (j + 3)'s cancel. and one of the (j + 4)'s in the top cancels with the (j + 4) in the bottom how? $\displaystyle $$\displaystyle \frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j-3)}{(j-4)(j+4)} $ it doesnt make sense beacuse there are 2 (j+3) at the bottom and only 1 (j+3) at the top...
Originally Posted by Jhevon yes all the (j + 3)'s cancel. and one of the (j + 4)'s in the top cancels with the (j + 4) in the bottom how? $\displaystyle \frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j-3)}{(j-4)(j+4)} $ it doesnt make sense beacuse there are 2 (j+3) at the bottom and only 1 (j+3) at the top...
Originally Posted by Ari how? $\displaystyle \frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j-3)}{(j-4)(j+4)} $ it doesnt make sense beacuse there are 2 (j+3) at the bottom and only 1 (j+3) at the top... there is a (j + 3) in the top of the first fraction as well as the top of the last fraction $\displaystyle \frac {{\color{red} (j+3)}(j+4)}{{\color {red}(j+3)(j+3)}} \cdot \frac {{\color{red}(j+3)}(j-3)}{(j-4)(j+4)}$
Originally Posted by Ari how? $\displaystyle \frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j-3)}{(j-4)(j+4)} $ it doesnt make sense beacuse there are 2 (j+3) at the bottom and only 1 (j+3) at the top... the (j - 3) in the top of the second fraction should be (j + 4)
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