Question

Fill in the blanks.
(i) $\left(\frac{-3}{17}\right)+\left(\frac{-12}{5}\right)=\left(\frac{-12}{5}\right)+\left(......\right)$
(ii) $-9+\frac{-21}{8}=\left(......\right)+\left(-9\right)$
(iii) $\left(\frac{-8}{13}+\frac{3}{7}\right)+\left(\frac{-13}{4}\right)=\left(......\right)+\left[\frac{3}{7}+\left(\frac{-13}{4}\right)\right]$
(iv) $-12+\left(\frac{7}{12}+\frac{-9}{11}\right)=\left(-12+\frac{7}{12}\right)+\left(......\right)$
(v) $\frac{19}{-5}+\left(\frac{-3}{11}+\frac{-7}{8}\right)=\left\{\frac{19}{-5}+\left(......\right)\right\}+\frac{-7}{8}$
(vi) $\frac{-16}{7}+......=......+\frac{-16}{7}=\frac{-16}{7}$

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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