(a) On the number line, it will mean four jumps of \(\frac{6}{5}\) to the left from zero. We reach at \(\frac{-24}{5}.\)

So, \(\frac{-6}{5}\times4=\frac{-24}{5}\)
(b) \(\frac{-6}{5}\times4=\frac{-6\times4}{5}=\frac{-24}{5}\)
We observe that we arrive at the same rational number.

(a) On the number line, it will mean three jumps of \(\frac{4}{7}\) to the left from zero. We reach at \(\frac{-12}{7}.\)

So, \(\frac{-4}{7}\times3=\frac{-12}{7}\)
(b) \(\frac{-4}{7}\times3=\frac{-4\times3}{7}=\frac{-12}{7}\)
We observe that we arrive at the same rational number.

\(\frac{7}{24}-\frac{17}{36}\)=\(\frac{7\times3-17\times2}{72}\)
=\(\frac{21-34}{72}=\frac{-13}{72}\)
(b) \(\frac{5}{63}-(\frac{-6}{21})\)=\(\frac{5}{63}+\frac{6}{21}=\frac{5+6+\times3}{63}\)
\(\frac{5+18}{63}=\frac{23}{63}\)
(c) \(\frac{-6}{13}-(\frac{-17}{15})\)=\(\frac{-6}{13}+\frac{7}{15}=\frac{-6\times15+7\times13}{195}\)
\(\frac{-90+91}{195}=\frac{1}{195}\)
(d) \(\frac{-3}{8}-\frac{7}{11}\)=\(\frac{-3\times11-7\times8}{88}\)
=\(\frac{-89}{88}=-1\frac{1}{88}\)
(e) \(-2\frac{1}{9}-6\)=\(\frac{-19}{9}-6=\frac{-19-6\times9}{9}\)
=\(\frac{-19-54}{9}=\frac{-73}{9}=-8\frac{1}{9}\)

(a) Clearly, the given rational numbers have a common and positive denominator. Arranging the numerators of given rational numbers\(\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}\) in ascending order, we get
-3<-2<-1
\(\Rightarrow \frac{-3}{5}<\frac{-2}{5}<\frac{-1}{5}\)
Hence, the given numbers when arranged in ascending order are \(\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}\)
(b)Clearly, the denominators of given rational numbers are positive. The denominators are 3, 9, 3. Their L.C.M. = 9. Writing the numbers so that they have a common denominator 9 as follows:
\(\frac{-1}{3}=\frac{-1\times3}{3\times3}=\frac{-3}{9},\frac{-2}{9}\)
and \(\frac{-4}{3}=\frac{-4\times3}{3\times3}=\frac{-12}{9}\)
Arranging the numerators of these rational numbers in ascending order, we get
-12 < -3 <-2
\(\Rightarrow \frac{-12}{9}<\frac{-3}{9}<\frac{-2}{9}\)
\(\Rightarrow \frac{-4}{3}<\frac{-1}{3}<\frac{-2}{9}\)
Hence, the given numbers when arranged in ascending order are \(\frac{-4}{3},\frac{-1}{3},\frac{-2}{9}\)
(c) Clearly, the denominators of given rational numbers are positive. The denominators are 7, 2, 4. Their L.C.M. is 28. Writing the numbers so that they have a common denominators 28as follows:
\(\frac{-3}{7}=\frac{-3\times4}{7\times4}=\frac{-12}{28}\)
\(\frac{-3}{2}=\frac{-3\times14}{2\times14}=\frac{-42}{28}\)
and, \(\frac{-3}{4}=\frac{-3\times7}{4\times7}=\frac{-21}{28}\)
or, - 42< - 21< - 12
or, \(\frac{-42}{28}=\frac{-21}{28}<\frac{-12}{28}\)
or, \(\frac{-3}{2}=\frac{-3}{4}<\frac{-3}{7}\)
Hence, the given numbers when arranged in ascending order are \(\frac{-3}{2},\frac{-3}{4},\frac{-3}{7}\)

x+y=\(\frac { -4 }{ 9 } +\frac { 5 }{ 12 } =\frac { -4\times 4+5\times 3 }{ 36 } =\frac { -16+15 }{ 36 } \)
\(\\ \Rightarrow\) x+y\(=\frac { -1 }{ 36 }\)
The reciprocal of x+y = \( \frac { 1 }{ \frac { -1 }{ 36 } } =-36\)

Suppose, if A is multiplied by y, then we get x
i.e.,  A\(\times \)y=x \(\\ \Rightarrow \) A\(\times \) \(\frac { 5 }{ 12 } \) = \(\\ \\ \frac { -4 }{ 9 } \)
A = \(A=\frac { -4 }{ 9 } \times \frac { 12 }{ 5 } =\frac { -48 }{ 45 } \)

0=\(\frac { 0 }{ 1 } \)

=\(\frac { 1 }{ 4 } \)

\(\\ 1\div \left( -\frac { 1 }{ 2 } \right) \)
The reciprocal of \(\left( \frac { -1 }{ 2 } \right) \) is \(\frac { 2 }{ -1\\ } \)
So,\(\\ \frac { 1 }{ 1 } \times \frac { 2 }{ -1 } =\frac { 1\times 2 }{ 1\times (-1) } =\frac { 2 }{ -1 } =-2\)

Given,\(\frac { 3 }{ 7 } \div \left( \frac { 21 }{ -55 } \right) \)
The reciprocal of  \(\\ \frac { 21 }{ -55 } \frac { -55 }{ 21 } \) .
So,\(\\ \frac { 3 }{ 7 } \times \frac { (-55) }{ 21 } =\frac { (-55)\times 3 }{ 7\times 21 } =\frac { -55 }{ 49 } \)