Question

How do you solve `(n-9)/(n+5)=7/4`?

Answer 1

See a solution process below:

Explanation:

First, cross multiply the equation:

`(color(orange)(n) - color(orange)(9))/(color(orange)(n) + color(orange)(5)) = color(blue)(7)/color(blue)(4)`

`color(blue)(4)(color(orange)(n) - color(orange)(9)) = color(blue)(7)(color(orange)(n) + color(orange)(5))`

`(color(blue)(4) * color(orange)(n)) - (color(blue)(4) * color(orange)(9)) = (color(blue)(7) * color(orange)(n)) + (color(blue)(7) * color(orange)(5))`

`4n - 36 = 7n + 35`

Next, subtract `color(red)(4n)` and `color(blue)(35)` from each side of the equation to isolate the `n` term while keeping the equation balanced:

#-color(red)(4n) + 4n - 36 - color(blue)(35) = -color(red)(4n) + 7n + 35 - color(blue)(35)#

`0 - 71 = (-color(red)(4) + 7)n + 0`

`-71 = 3n`

Now, divide each side of the equation by `color(red)(3)` to solve for `n` while keeping the equation balanced:

`-71/color(red)(3) = (3n)/color(red)(3)`

`-71/3 = (color(red)(cancel(color(black)(3)))n)/cancel(color(red)(3))`

`-71/3 = n`

`n = -71/3`

Answer 2

`n =-71/3`

Explanation:

In this equation there is one fraction on each side. The easiest method is to cross multiply. (See below for why it works).

`(n-9)/(n+5) = 7/4`

`7(n+5) = 4(n-9)`

`7n +35 = 4n -36`

`7n -4n = -36-35`

`3n = -71`

`n = -71/3`

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To get rid of the denominators, multiply both sides by the LCM of the denominators, which is `4(n+5)`

`(4(n+5))/1 xx (n-9)/(n+5) = (4(n+5))/1 xx 7/4" "larr` cancel

`(4cancel((n+5)))/1 xx (n-9)/cancel((n+5)) = (cancel4(n+5))/1 xx 7/cancel4`

This leaves you with: `4(n-9) = 7(n+5)`

Which is exactly the result from cross-multiplying.