How do you combine #(3z)/(z-3)-z/(z+4)#?

Question

1652 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-22T16:40:30

#{z(2z+15)}/{(z-3)(z+4)}#.

The Expression#=(3z)/(z-3)-z/(z+4)#

#={3z(z+4)-z(z-3)}/{(z-3)(z+4)}#

#={3z^2+12z-z^2+3z}/{(z-3)(z+4)}#

#=(2z^2+15z)/{(z-3)(z+4)}#

#={z(2z+15)}/{(z-3)(z+4)}#.

Answer 2 · 2016-08-22T16:41:05

#(3z(z+4)-z(z-3))/((z+4)(z-3))#

Or you can expand the brackets. I will let you do that

#color(blue)("Consider: "(3z)/(z-3)#

Multiply by 1 but in the form of #1=(z+4)/(z+4)# giving

#(3z)/(z-3)xx(z+4)/(z+4) = (3z(z+4))/((z-3)(z+4)#

'...........................................................................

#color(blue)("Consider: "-z/(z+4)#

Multiply by 1 but in the form #1=(z-3)/(z-3)#

#-z/(z+4)xx(z-3)/(z-3) = -(z(z-3))/((z+4)(z-3))#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Putting it all together")#

#(3z(z+4))/((z-3)(z+4)) -(z(z-3))/((z+4)(z-3))#

#(3z(z+4)-z(z-3))/((z+4)(z-3))#

Or you can expand the brackets. I will let you do that.