How do you simplify #\frac { 2z - 3} { z + 1} - \frac { 3} { z - 1} #?

Question

700 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-17T19:40:55

#(2z(z-4))/((z+1)(z-1))#

#color(blue)((2z-3)/(z+1))color(green)(-3/(z-1))#

You are subtracting 2 fractions, so:

first you need a common denominator....
In this case the denominator will be #(z+1)(z-1)#
make equivalent fractions.

#color(blue)((2z-3)/(z+1) xx (z-1)/(z-1) =((2z-3)(z-1))/((z+1)(z-1)))#

#color(green)(-3/(z-1)xx(z+1)/(z+1) = -(3(z+1))/((z+1)(z-1)) )#

Now add the numerators because the denominators are the same.

#((2z-3)(z-1))/((z+1)(z-1))-(3(z+1))/((z+1)(z-1)) #

#= ((2z-3)(z-1) -3(z+1))/((z+1)(z-1))#

#= (2z^2-2z-3z+3-3z-3)/((z+1)(z-1))#

#= (2z^2-8z)/((z+1)(z-1))" "larr# factorise the numerator

#= (2z(z-4))/((z+1)(z-1))#