Linear equation

[Guest]

(z-2)/(3-4z)+(z-3)/(4z-3)=(1)/(3)
Can someone help me with this equation?

Every time i tried to calculate it my answer was 0,25 but the answer is supposed to be 0.

[HallsofIvy]

HOW did you get 0.25= 1/4??

Clearly, if z= 1/4 then z- 2= 1/4- 8/4= -7/4, z- 3= 1/4- 12/4= -11/4, 3- 4z= 3- 1= 2 and 4z- 3= 1- 3= -2 (did you notice that 3- 4z= -(4z- 3)?) so this becomes
[tex]\frac{-7/4}{2}+ \frac{-11/4}{-2}= -\frac{7}{8}+ \frac{11}{8}= \frac{4}{8}= \frac{1}{2}[/tex], not [tex]\frac{1}{3}[/tex]. I wonder if you didn't accidently use 1/2 instead of 1/3?

[HallsofIvy]

First this is NOT a "linear equation", it is a "rational equation".

[tex]\frac{z- 2}{3- 4z}+ \frac{z- 3}{4z- 3}= \frac{1}{3}[/tex]

Get rid of the fraction by multiplying both sides by 3(4z- 3):
[tex]-3(z-2)+ 3(z- 3)= 4z- 3[/tex]

[tex]6- 9= -3= 4z- 3[/tex]. Can you solve that?