Question

How do you solve `6/(3b+2)=3/5`?

Answer 1

See the entire solution process below:

Explanation:

First, multiply each side of the equation by `color(red)(5)(color(blue)(3b + 2))` to eliminate the fraction while keeping the equation balanced:

`color(red)(5)(color(blue)(3b + 2)) xx 6/(3b + 2) = color(red)(5)(color(blue)(3b + 2)) xx 3/5`

`color(red)(5)cancel((color(blue)(3b + 2))) xx 6/color(blue)(cancel(color(black)((3b + 2)))) = cancel(color(red)(5))(color(blue)(3b + 2)) xx 3/color(red)(cancel(color(black)(5)))`

`5 xx 6 = (3b + 2) x 3`

`30 = (3b xx 3) + (2 xx 3)`

`30 = 9b + 6`

Next, subtract `color(red)(6)` from each side of the equation to isolate the `b` term while keeping the equation balanced:

`30 - color(red)(6) = 9b + 6 - color(red)(6)`

`24 = 9b + 0`

`24 = 9b`

Now, divide each side of the equation by `color(red)(9)` to solve for `b` while keeping the equation balanced:

`24/color(red)(9) = (9b)/color(red)(9)`

`(3 xx 8)/(3 xx 3) = (color(red)(cancel(color(black)(9)))b)/cancel(color(red)(9))`

`(cancel(3) xx 8)/(cancel(3) xx 3) = b`

`8/3 = b`

`b = 8/3`

Answer 2

`2 2/3`

Explanation:

`6/(3b+2)=3/5`

multiply both sides by `3b+2`

`:.cancel(3b+2)^1/1 xx 6/cancel(3b+2)^1=3/5 xx (3b+2)/1`

`:.1/1 xx6/1=(9b+6)/5`

`:.6=(9b+6)/5`

multiply both sides by `5/1`

`:.5/1 xx 6=(9b+6)/cancel5^1 xx cancel5^1/1`

`:.30=9b+6`

`:.9b+6=30`

`:.9b=30-6`

`:.9b=24`

`:.b=cancel24^8/cancel9^3`

`:.b=8/3` or `2 2/3`

substitute `x=8/3`

`:.6/(3(8/3)+2)=3/5`

`:.6/(cancel24^8/cancel33^1+2)=3/5`

`:.cancel6^3/cancel10^5=3/5`

`:.3/5=3/5`