Fill in the blanks.

Question:

Fill in the blanks.

(i) $\left(\frac{-3}{17}\right)+\left(\frac{-12}{5}\right)=\left(\frac{-12}{5}\right)+(\ldots \ldots)$

(ii) $-9+\frac{-21}{8}=(\ldots \ldots)+(-9)$

(iii) $\left(\frac{-8}{13}+\frac{3}{7}\right)+\left(\frac{-13}{4}\right)=(\ldots \ldots)+\left[\frac{3}{7}+\left(\frac{-13}{4}\right)\right]$

(iv) $-12+\left(\frac{7}{12}+\frac{-9}{11}\right)=\left(-12+\frac{7}{12}\right)+(\ldots \ldots)$

(v) $\frac{19}{-5}+\left(\frac{-3}{11}+\frac{-7}{8}\right)=\left\{\frac{19}{-5}+(\ldots \ldots)\right\}+\frac{-7}{8}$

(vi) $\frac{-16}{7}+\ldots \ldots=\ldots \ldots+\frac{-16}{7}=\frac{-16}{7}$

 

 

 

Solution:

(i) Addition is commutative, that is, $a+b=b+a$.

Hence, the required solution is $\left(\frac{-3}{17}\right)+\left(\frac{-12}{5}\right)=\left(\frac{-12}{5}\right)+\left(\frac{-3}{7}\right)$.

(ii) Addition is commutative, that is, $a+b=b+a$.

Hence, the required solution is $-9+\frac{-21}{8}=\frac{-21}{8}+-9$.

(iii) Addition is associative, that is, $(a+b)+c=a+(b+c)$.

Hence, the required solution is $\left(\frac{-8}{13}+\frac{3}{7}\right)+\left(\frac{-13}{4}\right)=\left(\frac{-8}{13}\right)+\left[\frac{3}{7}+\left(\frac{-13}{4}\right)\right]$.

(iv) Addition is associative, that is, $(a+b)+c=a+(b+c)$.

Hence, the required solution is $-12+\left(\frac{7}{12}+\frac{-9}{11}\right)=\left(-12+\frac{7}{12}\right)+\frac{-9}{11}$.

(v) Addition is associative, that is, $(a+b)+c=a+(b+c)$.

Hence, the required solution is $\frac{19}{-5}+\left(\frac{-3}{11}+\frac{-7}{8}\right)=\left\{\frac{19}{-5}+\left(\frac{-3}{11}\right)\right\}+\frac{-7}{8}$.

(vi) 0 is the additive identity, that is, $0+a=a$.

Hence, the required solution is $\frac{-16}{7}+0=0+\frac{-16}{7}=\frac{-16}{7}$.

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now