How do you simplify #(4z)/(z - 4) + (z + 4)/(z + 1)#?

Question

1607 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-05T15:51:20

#=(5z^2+4z-16)/((z-4)(z+1))#

#(4z)/(z-4) + (z+4)/(z+1)" "larr# find the LCD

Make equivalent fractions:

#=(4z)/(z-4) xx color(red)((z+1)/(z+1)) + (z+4)/(z+1) xx color(red)((z-4)/(z-4))#

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

#=(4z^2+4z+z^2-16)/((z-4)(z+1))#

#=(5z^2+4z-16)/((z-4)(z+1))#

NOTE:

#color(red)((z-4)/(z-4)) and color(red)((z+1)/(z+1))# are both equal to #1#.

Multiplying by #1# does not change the value.