Question 145604
{{{(((6r^2-7r-3)/(r^2-1)) /((4r^2-12r+9)/(r^2-1))) * ((2r^2-r-3)/(3r^2-2r-1))}}}
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Invert the second fraction in the first parentheses
and multiply:

{{{(((6r^2-7r-3)/(r^2-1))((r^2-1)/(4r^2-12r+9))) * ((2r^2-r-3)/(3r^2-2r-1))}}}

Cancel the {{{r^2-1}}}'s

{{{(((6r^2-7r-3)/cross(r^2-1))(cross(r^2-1)/(4r^2-12r+9))) * ((2r^2-r-3)/(3r^2-2r-1))}}}

{{{((6r^2-7r-3)/(4r^2-12r+9)) * ((2r^2-r-3)/(3r^2-2r-1))}}}

Factor all the numerators and denominators:

{{{(((2r-3)(3r+1))/((2r-3)(2r-3)))(((2r-3)(r+1))/(3r+1)(r-1)))  }}}

Cancel the {{{(2r-3)}}}'s

{{{((cross((2r-3))(3r+1))/(cross((2r-3))(2r-3)))(((2r-3)(r+1))/(3r+1)(r-1)))  }}}

Cancel the {{{(3r+1)}}}'s

{{{((cross((2r-3))cross((3r+1)))/(cross((2r-3))(2r-3)))(((2r-3)(r+1))/cross((3r+1))(r-1)))  }}} 

Cancel the {{{(2r-3)}}}'s

{{{((cross((2r-3))cross((3r+1)))/(cross((2r-3))cross((2r-3))))((cross((2r-3))(r+1))/cross((3r+1))(r-1)))  }}} 

{{{(r+1)/(r-1)}}}, which is choice B)

Edwin</pre>