Question 59829
Solve each equation. If a solution is extraneous, so indicate.
(5)/(2z+z-3)***-(2)/(2z+3)=(z+1)/(z-1)-(1)
***I think you meant to type (2z^2+z-3)
{{{5/(2z^2+z-3)-2/(2z+3)=(z+1)/(z-1)-1}}}  Factor the denominators.
{{{5/((2z+3)(z-1))-2/(2z+3)=(z+1)/(z-1)-1}}}
The LCD is(2z+3)(z-1), if you get a restriced value as an answer, it's extraneous. 
2z+3=0--->2z=-3--->z=-3/2 Is a restricted value that will give you a 0 in the denominator.
z-1=0-->z=1  Is also a restricted value.
Multiply everything by the LCD:
{{{5(2z+3)(z-1)/((2z+3)(z-1))-2(2z+3)(z-1)/(2z+3)=(z+1)(2z+3)(z-1)/(z-1)-1(2z+3)(z-1)}}}
{{{5*cross((2z+3)(z-1))/(cross((2z+3)(z-1)))-2*cross((2z+3))(z-1)/cross((2z+3))=(z+1)(2z+3)*cross((z-1))/cross((z-1))-1(2z+3)(z-1)}}}
{{{5-2(z-1)=(z+1)(2z+3)-1(2z+3)(z-1)}}}
{{{5-2z+2=2z^2+3z+2z+3-2z^2-z+3}}}
{{{(5+2)-2z=(2-2)z^2+(3+2-1)z+(3+3)}}}
{{{7-2z=4z+6}}}
{{{7-6-2z+2z=4z+2z+6-6}}}
{{{1=6z}}}
{{{1/6=z}}}
That's not a restrted value so the solutions is:{{{highlight(z=1/6)}}}
Happy Calculating!!!