# How do you solve 5/(b-5)=3/b?

Jul 10, 2016

$b = - \frac{15}{2}$

#### Explanation:

First we cross multiply and get:
$3 \left(b - 5\right) = 5 b$

Then we distribute the 3 on the left side
$3 b - 15 = 5 b$

Now we collect terms by subtracting $3 b$ from both sides.
$- 15 = 2 b$

Divide both sides by 2.
$- \frac{15}{2} = b$ which is our answer.

Jul 10, 2016

$b = - \frac{15}{2.}$

#### Explanation:

$\frac{5}{b - 5} = \frac{3}{b} .$
Multiplying by $b \left(b - 5\right)$ on both sides, we get,
$b \left(b - 5\right) \cdot \frac{5}{b - 5} = b \left(b - 5\right) \cdot \frac{3}{b}$
$\therefore \frac{5 b \left(b - 5\right)}{b - 5} = \frac{3 b \left(b - 5\right)}{b}$
$\therefore 5 b \frac{\cancel{b - 5}}{\cancel{b - 5}} = 3 \cancel{b} \frac{b - 5}{\cancel{b}}$
$\therefore 5 b = 3 \left(b - 5\right)$
$\therefore 5 b = 3 b - 15$
Adding, (-3b) on both sides, $5 b - 3 b = 3 b - 15 - 3 b$
$\therefore 2 b = \cancel{3} b - 15 - \cancel{3} b$
$\therefore 2 b = - 15$
Multiplying by $\frac{1}{2}$,
$\frac{1}{2} \left(2 b\right) = \frac{1}{2} \left(- 15\right)$
$\frac{1}{\cancel{2}} \cancel{2} b = - \frac{15}{2}$
$\therefore b = - \frac{15}{2.}$
This is the reqd. soln. Enjoy, Maths.!