SOLUTION: Solve each equation. If a solution is extraneous, so indicate. (5)/(2z+z-3)-(2)/(2z+3)=(z+1)/(z-1)-(1)

Question 59829This question is from textbook Intermediate Algebra
: Solve each equation. If a solution is extraneous, so indicate.
(5)/(2z+z-3)-(2)/(2z+3)=(z+1)/(z-1)-(1) This question is from textbook Intermediate Algebra

Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
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)
Factor the denominators.

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:

That's not a restrted value so the solutions is:
Happy Calculating!!!
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