Question

How do you simplify `\frac { k - 2} { 3k + 3} - \frac { k + 1} { 3k + 3}`?

Answer 1

`(k-2)/(3k+3)-(k+1)/(3k+3)=-1/(k+1)`

Explanation:

As in `(k-2)/(3k+3)` and `(k+1)/(3k+3)`,

denominator is common,

while adding such fractions we can add / subtract numerators keeping denominators same to get result.

Hence `(k-2)/(3k+3)-(k+1)/(3k+3)`

= `(k-2-(k+1))/(3k+3)`

= `(k-2-k-1)/(3k+3)`

= `(cancelk-2-cancelk-1)/(3k+3)`

= `-3/(3k+3)`

= `-(3xx1)/(3(k+1))`

= `-1/(k+1)`

Answer 2

the simplified term is `(-1)/(k+1)`

Explanation:

the given equation is `(k-2)/(3k+3)-(k+1)/(3k+3)` since the `LCM` is same we can directly subtract the given 2 terms
we get,`(k-2-k-1)/(3k+3)=(-3)/(3k+3)=-3/(3(k+1))=(-1)/(k+1)`