Question

How do you solve `\frac { w } { w - 3} - \frac { 5w } { 5w - 2} = \frac { w - 2} { 5w ^ { 2} - 17w + 6}`?

Answer 1

`w=-1/6`

Explanation:

First combine the fractions on the left hand side:

`w/(w-3)-(5w)/(5w-2)=(w-2)/(5w^2-17w+6)`

`(w(5w-2))/((5w-2)(w-3))-((5w)(w-3))/((5w-2)(w-3))=(w-2)/(5w^2-17w+6)`

`((5w^2-2w))/((5w-2)(w-3))-((5w^2-15w))/((5w-2)(w-3))=(w-2)/(5w^2-17w+6)`

`(13w)/(5w^2-17w+6)=(w-2)/(5w^2-17w+6)`

With the denominators equal, we can say:

`13w=w-2`

`12w=-2=>w=-1/6`

Now let's make sure that value isn't disallowed by the denominator. Remember that we had `w` terms in the denominator, so any value of `w` that results in a denominator of 0 can't be allowed.

`w!=3, 2/5`