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

2 Answers
Shwetank Mauria
Nov 26, 2017

#(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)#

Manasvini Nittala
Nov 26, 2017

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)#

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