# How do you solve (t+4)/t+3/(t-4)=-16/(t^2-4t)?

May 3, 2018

#### Answer:

$t = - 3$ is the sole solution.

#### Explanation:

We can assume ${t}^{2} - 4 t = t \left(t - 4\right) \ne 0 ,$ i.e. $t \setminus \ne 0 \mathmr{and} t \ne 4$ so we can cancel the common denominator when we get them.

$\frac{t + 4}{t} + \frac{3}{t - 4} = - \frac{16}{{t}^{2} - 4 t}$

$\frac{\left(t + 4\right) \left(t - 4\right) + 3 t}{t \left(t - 4\right)} = \frac{- 16}{t \left(t - 4\right)}$

$\left(t + 4\right) \left(t - 4\right) + 3 t = - 16$

${t}^{2} - 16 + 3 t = - 16$

$t \left(t + 3\right) = 0$

$t = 0$ was ruled out at the start

$t = - 3$ is the sole solution

Check:

$\frac{- 3 + 4}{-} 3 + \frac{3}{- 3 - 4} = - \frac{1}{3} - \frac{3}{7} = - \frac{16}{21}$

 -16/{(-3)^2 - 4(-3) } = -16/{21} quad sqrt