Question

How do you solve `3/(b+2)=4/b`?

Answer 1

See a solution process below:

Explanation:

First, because both sides of the equation are pure fractions we can flip each equation to rewrite the equation as:

`(b + 2)/3 = b/4`

Next, multiply each side of the equation by `color(red)(12)` to eliminate the fractions while keeping the equation balanced. `color(red)(12)` is used because it is the lowest common denominator of the two fractions:

`color(red)(12) xx (b + 2)/3 = color(red)(12) xx b/4`

`cancel(color(red)(12))color(red)(4) xx (b + 2)/color(red)(cancel(color(black)(3))) = cancel(color(red)(12))color(red)(3) xx b/color(red)(cancel(color(black)(4)))`

`color(red)(4)(b + 2) = 3b`

`(color(red)(4) xx b) + (color(red)(4) xx 2) = 3b`

`4b + 8 = 3b`

Now, subtract `color(red)(8)` and `color(blue)(3b)` from each side of the equation to solve for `b` while keeping the equation balanced:

`4b - color(blue)(3b) + 8 - color(red)(8) = 3b - color(blue)(3b) - color(red)(8)`

`(4 - color(blue)(3))b + 0 = 0 - color(red)(8)`

`1b = -8`

`b = -8`

Answer 2

I get `b=-8`

Explanation:

To solve equations with fractions use the rule of cross products. Each numerator is multiplied by the opposite denominator and the two products are equal.

`\frac{color(blue)(3)}{b+2}=\frac{4}{color(blue)(b)}`

`color(blue)(3b)=(4)(b+2)`

Now it's just an ordinary linear equation:

`3b=4b+8`

`-b=8`

`b=-8`

Check:

`\frac{3}{-8+2}=\frac{3}{-6}=-\frac{1}{2}`

`\frac{4}{-8}=-\frac{1}{2}`