Question

How do you solve `(n-7)/5+n/8=(n-4)/2+2`?

Answer 1

`n=-8`

Explanation:

Note: This is a long answer.

Solve `(n-7)/5+n/8=(n-4)/2+2`

In order to add and subtracting fractions, the denominators must be the same. To do this, we look for the least common denominator (LCD). One way to do this is by prime factorization. List the prime factors for each denominator.

LCD

Prime factorization of `5:` `5`
Prime factorization of `8:` `2xx2xx2`
Prime factorization of `2:` `2`

LCD: `2xx2xx2xx5=40`

Now multiply each fraction by an equivalent fraction that will make all three denominators equal to the LCD `40`.

`(n-7)/5(8/8)+n/8(5/5)=(n-4)/2(20/20)+2(40/40)`

Simplify.

`(8(n-7))/40+(5n)/40=(20(n-4))/40+80/40`

Simplify.

Multiply each fraction by `40`.

`cancel(40)((8(n-7))/cancel 40)+cancel 40((5n)/cancel 40)=cancel(cancel(40))((20(n-4))/cancel(40))+cancel(40)(80/cancel(40))`

Simplify.

`8(n-7)+5n=20(n-4)+80`

Expand both sides using the distributive property.

`8n-56+5n=20n-80+80`

Simplify.

`13n-56=20n`

Add `56` to both sides.

`13n=20n+56`

Subtract `20n` from both sides.

`-7n=56`

Divide both sides by `-7`.

`n=(56)/(-7)`

Simplify.

`n=-56/7`

Simplify.

`n=-8`