Question

How do you solve `\frac { t } { t - 3} + \frac { 4} { t + 3} = \frac { 18} { t ^ { 2} - 9}`?

Answer 1

`t=-10,3`

Explanation:

Multiply both sides by `t^2-9`:

`t(t+3)+4(t-3)=18`

Now distribute:

`t^2+3t+4t-12=18`

Combine like terms:

`t^2+7t-12=18`

Move the 18 to the other side:

`t^2+7t-30=0`

Factor:

`(t+10)(t-3)=0`

Solve for t:

`t=-10,3`

Answer 2

`t=3` and `t=-10`

Explanation:

This is a long process, so let's take this one step at a time.
First, let's see what we can do about those denominators
`t/(t-3)+4/(t+3)=18/(t^2-9)`. Now, I noticed that if we multiply `(t-3)` by `(t+3)`, we'd get `t^2-9`, which is the same as the denominator on the right. Let's multiply the fractions to make the denominators `t^2-9`.

`(t)/(t-3)*((color(green)(t+3))/(color(green)(t+3)))` `+4/(t+3)` `((color(green)(t-3))/(color(green)(t-3)))=18/(t^2-9)`

This might look ugly, but let's simplify a bit:
`(t^2+3t)/(t^2-9)+(4t-12)/(t^2-9)=18/(t^2-9)`
Since the denominators are now the same,l we can combine them. That gives us `(t^2+3t+4t-12)/(t^2-9)=18/(t^2-9)`. If we multiply both sides by `t^2-9`, that should get rid of the denominators. So now we have `(t^2+3t+4t-12)/cancel(t^2-9)=18/cancel(t^2-9)*cancel(t^2-9)` or `t^2+3t+4t-12=18`. Now we just need to combine like-terms and factor.

`t^2+7t-30=0`
We are looking for two number that multiply to `30` and add to `-7`. Those numbers are `-3` and `10`. So, the factors are `(t-3)(t+10)`

We just set both these factors to `0` and solve fot `t`.
`t-3=0` `color(white)(..............)` `t+10=0`
`t=3` `color(white)(......000........0)` `t=-10`