Question 6733
So This is probably {{{ (3-x)/((3x-1)(x+1)) = 0 }}} Maybe it'd help if you rewrote it as


{{{ (3-x)/((3x-1)(x+1)) = 0/1 }}} <---- and then cross-multiply. You'll then get


{{{ 3 - x = 0 }}} <--- just by looking at this equation, x = 3.


It seems almost stupid, but if you plug in 3 into the equation, you'll get a zero in the numerator. It won't matter what the denominator is because {{{ 0/n = 0 }}} just as long as that denominator n doesn't turn out to be 0.