RD Sharma Solutions Class 8 Rational Numbers Exercise 1.2

RD Sharma Solutions Class 8 Chapter 1 Exercise 1.2

RD Sharma Class 8 Solutions Chapter 1 Ex 1.2 PDF Free Download

Exercise 1.2

Q-1. Verify commutativity of addition of rational numbers for each of the following pairs of rational numbers.

(i)  \(\frac{ -11 }{ 5 }\) and \(\frac{ 4 }{ 7 }\)

(ii)  \(\frac{ 4 }{ 9 }\) and \(\frac{ 7 }{ -12 }\)

(iii)  \(\frac{ -3 }{ 5 }\) and \(\frac{ -2 }{ -15 }\)

(iv)  \(\frac{ 2 }{ -7 }\) and \(\frac{ 12 }{ -35 }\)

(v)  4 and \(\frac{ -3 }{ 5 }\)

(vi)  -4 and \(\frac{ 4 }{ -7 }\)

 

Solution:

Commutativity of the addition of rational numbers means that if ab and cd are two rational numbers, then ab + cd = cd + ab.

(i) We have:

\(\frac{ -11 }{ 5 }\) and \(\frac{ 4 }{ 7 }\)

So,

\(\frac{ -11 }{ 5 }\)\(\frac{ 4 }{ 7 }\)

= \(\frac{ -11 \times 7 }{ 5 \times 7 }\) + \(\frac{ 4 \times 5  }{ 7 \times 5}\)

= \(\frac{ -77 }{ 35 }\) + \(\frac{ 20 }{ 35 }\)

\(\frac{ -77 + 20 }{ 35 }\)

\(\frac{ -57 }{ 35 }\)

Now,

\(\frac{ 4 }{ 7 }\)  and \(\frac{ -11 }{ 5 }\)

= \(\frac{ 4 }{ 7 }\) + \(\frac{ -11 }{ 5 }\)

= \(\frac{ 4 \times 5 }{ 7 \times 5 }\) + \(\frac{ -11 \times 7  }{ 5 \times 7 }\)

= \(\frac{ 20 }{ 35 }\) and \(\frac{ -77 }{ 35}\)

= \(\frac{ 20 – 77 }{ 35 }\) = \(\frac{ -57 }{ 35 }\)

Hence, verified.

(ii)  We have:

\(\frac{ 4 }{ 9 }\) and \(\frac{ 7 }{ -12 }\)

So,

\(\frac{ 4  }{ 9 }\) + \(\frac{ -7 }{ 12 }\)

= \(\frac{ 4 \times 4 }{ 9 \times 4 }\) + \(\frac{ -7 \times 3 }{ 12 \times 3 }\)

= \(\frac{ 16 }{ 36 }\) + \(\frac{ -21 }{ 36 }\)

= \(\frac{ 16 – 21 }{ 36 }\)

= \(\frac{ -5 }{ 36 }\)

Now,

\(\frac{ -7 }{ 12 }\) and \(\frac{ 4 }{ 9 }\)

= \(\frac{ -7 }{ 12 }\) + \(\frac{ 4 }{ 9 }\)

= \(\frac{ -7 \times  3 }{ 12 \times 3 }\) + \(\frac{ 4 \times 4 }{ 9 \times 4 }\)

= \(\frac{ -21 }{ 36 }\) + \(\frac{ 16 }{ 36 }\)

= \(\frac{ -21 + 16 }{ 36 }\) = \(\frac{ -5 }{ 36 }\)

Hence, Verified.

(iii)  We have:

\(\frac{ -3 }{ 5 }\) and \(\frac{ -2 }{ -15 }\)

So,

\(\frac{ -3 }{ 5 }\) + \(\frac{ 2 }{ 15 }\)

= \(\frac{ -3 \times 3 }{ 5 \times 3  }\) +\(\frac{ 2 }{ 15 }\)

= \(\frac{ -9 }{ 15 }\) + \(\frac{ 2 }{ 15 }\)

= \(\frac{ -9 + 2 }{ 15 }\) = \(\frac{ -7 }{ 15 }\)

Now,

\(\frac{ 2 }{ 15 }\) and \(\frac{ -3  }{ 5 }\)

= \(\frac{ 2  }{ 15 }\) + \(\frac{ -3 }{ 5 }\)

= \(\frac{ 2 }{ 15 }\) + \(\frac{ -3 \times 3 }{ 5 \times 3 }\)

= \(\frac{ 2 }{ 15 }\) + \(\frac{ -9 }{ 15 }\)

= \(\frac{ 2 – 9 }{ 15 }\) = \(\frac{ -7 }{ 15 }\)

Hence, verified.

(iv) We have:

\(\frac{ 2 }{ -7 }\) and \(\frac{ 12 }{ -35 }\)

So,

\(\frac{ -2 }{ 7 }\) + \(\frac{ -12 }{ 35 }\)

= \(\frac{ -2 \times 5 }{ 7 \times 5 }\) + \(\frac{ -12 }{ 35 }\)

= \(\frac{ -10 }{ 35 }\) + \(\frac{ -12 }{ 35 }\)

= \(\frac{ -10 – 12 }{ 35 }\)

= \(\frac{ -22 }{ 35 }\)

Now,

\(\frac{ -12 }{ 35 }\) and \(\frac{ -2 }{ 7 }\)

= \(\frac{ -12 }{ 35 }\) + \(\frac{ -2 \times 5 }{ 7 \times 5 }\)

= \(\frac{ -12 }{ 35 }\) + \(\frac{ -10 }{ 35 }\)

= \(\frac{ -12 – 10 }{ 35 }\) = \(\frac{ -22 }{ 35 }\)

Hence, verified.

(v)  We have:

\(\frac{ 4 }{ 1 }\) and \(\frac{ -3 }{ 5 }\)

So,

\(\frac{ 4 }{ 1 }\) + \(\frac{ -3 }{ 5 }\)

= \(\frac{ 4 \times  5 }{ 1 \times  5 }\) + \(\frac{ -3 }{ 5 }\)

= \(\frac{ 20 }{ 5 }\) + \(\frac{ -3 }{ 5 }\)

= \(\frac{ 20 – 3 }{ 5 }\) = \(\frac{ 17 }{ 5 }\)

Now,

\(\frac{ -3 }{ 5 }\) and \(\frac{ 4 }{ 1 }\)

= \(\frac{ -3 }{ 5 }\) + \(\frac{ 4 }{ 1 }\)

= \(\frac{ -3 }{ 5 }\) + \(\frac{ 4 \times 5 }{ 1 \times 5 }\)

= \(\frac{ -3 }{ 5 }\) + \(\frac{ 20 }{ 5 }\)

= \(\frac{ -3 + 20 }{ 5 }\) = \(\frac{ -17 }{ 5 }\)

Hence, verified.

(vi) We have:

\(\frac{ -4 }{ 1 }\) and \(\frac{ 4 }{ -7 }\)

So,

\(\frac{ -4 }{ 1 }\) + \(\frac{ -4 }{ 7 }\)

= \(\frac{ – 4 \times 7 }{ 1 \times  7 }\) + \(\frac{ -4 }{ 7 }\)

= \(\frac{ -28 }{ 7 }\) + \(\frac{ -4 }{ 7 }\)

= \(\frac{-28 – 7 }{ 7 }\) = \(\frac{ -35 }{ 7 }\) = 5

Now,

\(\frac{ -4 }{ 7 }\) and  \(\frac{ -4 }{ 1 }\)

= \(\frac{ -4 }{ 7 }\) + \(\frac{ -4 }{ 1 }\)

= \(\frac{ -4 }{ 7 }\) + \(\frac{ -4 \times 7 }{ 1 \times 7 }\)

= \(\frac{ -4 }{ 7 }\) + \(\frac{ -28 }{ 7 }\)

= \(\frac{ -4 – 28 }{ 7 }\) = \(\frac{ -35 }{ 7 }\)

Hence, verified.

Q-2. Verify associativity of addition of the rational numbers i.e., \(\left ( x + y \right ) + z = x + \left ( y + z \right )\), when:

(i)   \(x = \frac{ 1 }{ 2 } , y = \frac{ 2 }{ 3 } , z = -\frac{ 1 }{ 5 }\)

(ii)  \(x = \frac{- 2 }{ 5 } , y = \frac{ 4 }{ 3 } , z = -\frac{ 7 }{ 10 }\)

(iii) \(x = – \frac{ 7 }{ 11 } , y = -\frac{ 2 }{ 5 } , z = -\frac{ 3 }{ 22 }\)

(iv) \(x = -2 , y = \frac{ 3 }{ 5 } , z = -\frac{ 4 }{ 3 }\)

 

Solution:

We have to verify that:

\(\left ( x + y \right ) + z = x + \left ( y + z \right )\)

(i) \(x = \frac{1}{2} , y = \frac{2}{3} , z = -\frac{1}{5}\)

= \(\left ( \frac{1}{2} + \frac{2}{3} \right ) + \left (-\frac{1}{5} \right )\)

= \(\left ( \frac{7}{6} \right ) – \frac{1}{5}\)

= \(\left ( \frac{7 \times 5}{6 \times 5 } \right ) – \frac{1 \times 7 }{5 \times  7 }\)

= \(\left ( \frac{ 35 }{ 30 } \right ) – \frac{ 7 }{ 35 }\)

= \(\frac{ 35 – 7 }{ 30 }\)

= \(\frac{ 29 }{ 30 }\)

Now,

\(\frac{1}{2} + \left ( \frac{ 2 }{ 3 } + \frac{ -1 }{ 5 }\right )\)

= \(\frac{1}{2} + \left ( \frac{ 2 \times 5 }{ 3 \times  5 } + \frac{ -1 \times 3 }{ 5 \times 3 }\right )\)

= \(\frac{1}{2} + \left ( \frac{ 10 }{ 15 } + \frac{ -3 }{ 15 }\right )\)

= \(\frac{1}{2} + \left ( \frac{ 10 – 3 }{ 15 }  \right )\)

= \(\frac{1}{2} + \left ( \frac{ 7 }{ 15 } \right )\)

= \(\frac{ 1 \times 15 }{ 2 \times 15 } + \frac{ 7 \times 2}{15 \times 2}\)

= \(\frac{ 15 }{ 30 } + \frac{ 14}{ 30 }\)

= \(\frac{ 15 + 14 }{ 30 }\) = \(\frac{ 29 }{ 30 }\)

Hence, verified.

(ii)  \(x = \frac{- 2 }{ 5 } , y = \frac{ 4 }{ 3 } , z = -\frac{ 7 }{ 10 }\)

= \(\left (\frac{- 2 }{ 5 } + \frac{ 4 }{ 3 } \right ) -\frac{ 7 }{ 10 }\)

= \(\left (\frac{- 2 \times 3  }{ 5 \times 3 } + \frac{ 4 \times  5 }{ 3 \times 5 } \right ) -\frac{ 7 }{ 10 }\)

= \(\left (\frac{- 6 }{ 15 } + \frac{ 20 }{ 15 } \right ) -\frac{ 7 }{ 10 }\)

= \(\left (\frac{- 6 + 20 }{ 15 } \right ) -\frac{ 7 }{ 10 }\)

= \(\left (\frac{ 14 }{ 15 } \right ) -\frac{ 7 }{ 10 }\)

= \(\frac{ 14 \times 2 }{ 15 \times 2 } – \frac{ 7 \times 3 }{ 10 \times 3 }\)

= \(\frac{ 28 }{ 30 } – \frac{ 21 }{ 30 }\)

= \(\frac{ 28 – 21 }{ 30 }\) = \(\frac{ 7 }{ 30 }\)

Now,

\(\frac{- 2 }{ 5 } + \left (\frac{ 4 }{ 3 } -\frac{ 7 }{ 10 } \right )\)

= \(\frac{- 2 }{ 5 } + \left (\frac{ 4 \times 10 }{ 3 \times 10 } -\frac{ 7 \times 3 }{ 10 \times  3} \right )\)

= \(\frac{- 2 }{ 5 } + \left (\frac{ 40 }{ 30 } -\frac{ 21 }{ 30 } \right )\)

= \(\frac{- 2 }{ 5 } + \left (\frac{ 40 – 21  }{ 30 } \right )\)

= \(\frac{- 2 }{ 5 } + \left (\frac{ 19 }{ 30 } \right )\)

= \(\frac{- 2 \times 6 }{ 5 \times 6  } + \left (\frac{ 19 }{ 30 } \right )\)

= \(\frac{- 12 }{ 30 } + \left (\frac{ 19 }{ 30 } \right )\)

= \(\frac{- 12 + 19 }{ 30 }\) = \(\frac{ 7 }{ 30 }\)

Hence, verified.

(iii) x = –\(-\frac{ 7 }{9 }\), y =\( \frac{ 2 }{ -5 }\) , z = \(-\frac{ 3 }{ 22 }\)

= \(\left (\frac{- 7 }{ 11 } + \frac{ 2 }{ -5 } \right ) -\frac{ 3 }{ 22 }\)

= \(\left (\frac{- 7 }{ 11 } + \frac{ -2 }{ 5 } \right ) -\frac{ 3 }{ 22 }\)

= \(\left (\frac{- 7 \times 5 }{ 11 \times 5 } + \frac{ -2 \times 11 }{ 5 \times 11 } \right ) -\frac{ 3 }{ 22 }\)

= \(\left (\frac{- 35 }{ 55 } + \frac{ -22 }{ 55 } \right ) -\frac{ 3 }{ 22 }\)

= \(\left (\frac{- 35 – 22 }{ 55 } \right ) -\frac{ 3 }{ 22 }\)

= \(\left (\frac{- 57 }{ 55 } \right ) -\frac{ 3 }{ 22 }\)

= \(\frac{-57 \times 2}{ 55 \times 2 } -\frac{ 3 \times 5}{ 22 \times 5}\)

= \(\frac{-114}{ 110 } -\frac{ 15}{ 110 }\)

= \(\frac{-114 – 15 }{ 110 }\) = \(\frac{-129 }{ 110 }\)

Now,

\(\frac{- 7 }{ 11 } + \left (\frac{ -2 }{ 5 } – \frac{ 3 }{ 22 } \right )\)

= \(\frac{- 7 }{ 11 } + \left (\frac{ -2 \times 22 }{ 5 \times 22 } – \frac{ 3 \times 5}{ 22 \times 5 } \right )\)

= \(\frac{- 7 }{ 11 } + \left (\frac{ -44 }{ 110 } – \frac{ 15 }{ 110 } \right )\)

= \(\frac{- 7 }{ 11 } + \left (\frac{ -44 – 15 }{ 110 } \right )\)

= \(\frac{- 7 }{ 11 } + \left (\frac{ -59 }{ 110 } \right )\)

= \(\frac{- 7 \times 10 }{ 11  \times 10 } + \left (\frac{ -59 }{ 110 } \right )\)

= \(\frac{- 70 }{ 110 } + \frac{ -59 }{ 110 } \) = \(\frac{-129 }{ 110 }\)

Hence, verified.

(iv) \(x = -2 , y = \frac{ 3 }{ 5 } , z = -\frac{ 4 }{ 3 }\)

= \(\left ( -2 + \frac{ 3 }{ 5 } \right ) -\frac{ 4 }{ 3 }\)

= \(\left ( -2 \times 5 + \frac{ 3 }{ 5 } \right ) -\frac{ 4 }{ 3 }\)

= \(\left (\frac{ -10 + 3 }{ 5 } \right ) -\frac{ 4 }{ 3 }\)

= \(\frac{ -7 }{ 5 } -\frac{ 4 }{ 3 }\)

= \(\frac{ -7 \times 3 }{ 5 \times 3} -\frac{ 4 \times 5 }{ 3 \times 5 }\)

= \(\frac{ -21 }{ 15 } -\frac{ 20 }{ 1 5 }\)

= \(\frac{ -21 – 20 }{ 15 }\) = \(\frac{ -41 }{ 15 }\)

Now,

\(-2 + \left (\frac{ 3 }{ 5 } -\frac{ 4 }{ 3 } \right )\)

= \(-2 + \left (\frac{ 3 \times 3}{ 5 \times 3 } -\frac{ 4 \times 5 }{ 3 \times 5 } \right )\)

= \(-2 + \left (\frac{ 9 }{ 15 } -\frac{ 20 }{ 15 } \right )\)

= \(-2 + \left (\frac{ 9 – 20}{ 15 } \right )\)

= \(-2 + \left (\frac{ -11 }{ 15 } \right )\)

= \(-\frac{2 \times 15}{ 1 \times 15 } + \frac{ – 11 }{ 15 }\)

= \(-\frac{ 30 }{ 15 } + \frac{ – 11 }{ 15 }\)

= \(\frac{ -30 – 11 }{ 15 }\) = \(\frac{ -41 }{ 15 }\)

Hence, verified.

Q-3. Write the additive inverse of each of the following rational numbers:

(i) \(\frac{ -2 }{ 17 }\)

(ii) \(\frac{ 3 }{ -11 }\)

(iii) \(\frac{ -17 }{ 5  }\)

(iv) \(\frac{ -11  }{ -25 }\)

Solution:

(i) Additive inverse is the negative of the given number.

So, additive inverse of \(\frac{ -2 }{ 17 }\) = \(\frac{ 2 }{ 17 }\)

(ii) Additive inverse is the negative of the given number.

So, additive inverse of \(\frac{ 3 }{ -11 }\) = \(\frac{ 3 }{ 11 }\)

(iii) Additive inverse is the negative of the given number.

So, additive inverse of \(\frac{ -17 }{ 5 }\) = \(\frac{ 17 }{ 5 }\)

(iv) Additive inverse is the negative of the given number.

So, additive inverse of \(\frac{ -11 }{ -25 }\) = \(\frac{ -11 }{ 25 }\)

Q-4. Write the negative (additive inverse) of each of the following:

(i) \(\frac{ -2 }{ 5 }\)

(ii) \(\frac{ 7  }{ -9 }\)

(iii) \(\frac{ -16 }{ 13 }\)

(iv) \(\frac{ -5 }{ 1 }\)

(v) 0

(vi) 1

(vii) -1

 

Solution:

(i) Additive inverse of \(\frac{ -2 }{ 5 }\) = \(\frac{ 2 }{ 5 }\)

(ii) Additive inverse of \(\frac{ -7 }{ 9 }\) = \(\frac{ 7 }{ 9 }\)

(iii) Additive inverse of \(\frac{ -16 }{ 13 }\) = \(\frac{ 16 }{ 13 }\)

(iv) Additive inverse of \(\frac{ -5 }{ 1 }\) = \(\frac{ 5 }{ 1 }\)

(v) Negative value of 0 is 0

(vi) Negative value of 1 is -1

(vii) Negative value of -1 is 1

Q-5. Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:

(i)  \(\frac{ 2 }{ 5 } + \frac{ 7 }{ 3 } + \frac{ -4 }{ 5 } + \frac{ -1 }{ 3 }\)

(ii)  \(\frac{ 3 }{ 7 } + \frac{ -4 }{ 9 } + \frac{ -11 }{ 7 } + \frac{ 7 }{ 9 }\)

(iii)  \(\frac{ 2 }{ 5 } + \frac{ 8 }{ 3 } + \frac{ -11 }{ 15 } + \frac{ 4 }{ 5 } +  \frac{ -2 }{ 3 }\)

(iv)  \(\frac{ 4 }{ 7 } + 0 + \frac{ -8 }{ 9 } + \frac{ -13 }{ 7 } + \frac{ 17 }{ 21 }\)

Solution:

(i)  We have:

\(\frac{ 2 }{ 5 } + \frac{ 7 }{ 3 } + \frac{ -4 }{ 5 } + \frac{ -1 }{ 3 }\)

= \(\left (\frac{ 2 }{ 5 } + \frac{ -4 }{ 5 } \right ) + \left ( \frac{ -1 }{ 3 } + \frac{ 7 }{ 3 } \right )\)

= \(\frac{ 2 – 4 }{ 5 } + \frac{ -1 + 7 }{ 3 }\)

= \(\frac{ -2 }{ 5 } + \frac{ 6 }{ 3 }\)

= \(\frac{ -2 \times 3 }{ 5 \times 3 } + \frac{ 6 \times 5 }{ 3 \times 5 }\)

= \(\frac{ -6 }{ 15 } + \frac{ 30 }{ 15 }\)

= \(\frac{ -6 + 30 }{ 15 } \) = \(\frac{ 24 }{ 15 }\) = \(\frac{ 8 }{ 5 }\)

(ii) \(\frac{ 3 }{ 7 } + \frac{ -4 }{ 9 } + \frac{ -11 }{ 7 } + \frac{ 7 }{ 9 }\)

= \(\left (\frac{ 3 }{ 7 } + \frac{ -11 }{ 7 } \right ) + \left ( \frac{ -4 }{ 9 } + \frac{ 7 }{ 9 } \right )\)

= \(\frac{ 3 – 11 }{ 7 } + \frac{ -4 + 7 }{ 9 }\)

= \(\frac{ -8 }{ 7 } + \frac{ 3 }{ 9 }\)

= \(\frac{ -8 \times 9 }{ 7 \times 9 } + \frac{ 3 \times 7 }{ 9 \times 7 }\)

= \(\frac{ -72 }{ 63 } + \frac{ 21 }{ 63 }\)

= \(\frac{ -72 + 21 }{ 63 } \)

= \(\frac{ -51 }{ 63 }\) = \(\frac{ -17 }{ 21 }\)

(iii) \(\frac{ 2 }{ 5 } + \frac{ 8 }{ 3 } + \frac{ -11 }{ 15 } + \frac{ 4 }{ 5 } + \frac{ -2 }{ 3 }\)

= \(\left (\frac{ 2 }{ 5 } + \frac{ 8 }{ 3 } \right ) + \left ( \frac{ -11 }{ 15 } + \frac{ 4 }{ 5 }  + \frac{ -2 }{ 3 } \right )\)

= \(\frac{ 2 + 4 }{ 5 } + \frac{ 8 -2  }{ 3 } + \frac{ -11 }{ 15 }\)

= \(\frac{ 6 }{ 5 } + \frac{ 6 }{ 3 } + \frac{ -11 }{ 15 }\)

= \(\frac{ 6 \times 3 }{ 5 \times 3 } + \frac{ 6 \times 5 }{ 3 \times 5 } + \frac{ -11 }{ 15 }\)

= \(\frac{ 18 }{ 15 } + \frac{ 30 }{ 15 } + \frac{ -11 }{ 15 }\)

= \(\frac{ 18 + 30 }{ 15 } + \frac{ -11 }{ 15 }\)

= \(\frac{ 48 }{ 15 } + \frac{ -11 }{ 15 }\)

= \(\frac{ 48 – 11 }{ 15 }\) = \(\frac{ 37 }{ 15 }\)

(iv) \(\frac{ 4 }{ 7 } + 0 + \frac{ -8 }{ 9 } + \frac{ -13 }{ 7 } + \frac{ 17 }{ 21 }\)

= \(\left (\frac{ 4 }{ 7 } + \frac{ -13 }{ 7 } \right ) + \frac{ -8 }{ 9 } + \frac{ 17 }{ 21 } \)

= \(\frac{ 4 – 13 }{ 7 } + \frac{ -8 }{ 9 } + \frac{ 17 }{ 21 }\)

= \(\frac{ -9 }{ 7 } + \frac{ -8 }{ 9 } + \frac { 17 }{ 21 }\)

= \(\frac{ -9 \times 9 }{ 7 \times 9 } + \frac{ -8 \times 7 }{ 9 \times 7 } + \frac { 17 \times  3 }{ 21 \times 3 }\)

= \(\frac{ -81 }{ 63 } + \frac{ -56 }{ 63 } + \frac{ 51 }{ 63 }\)

= \(\frac{ -81 – 56 + 51 }{ 63 } \)

= \(\frac{ -86 }{ 63 } \)

Q-6. Re-arrange suitably and find the sum of each of the following:

(i) \(\frac{ 11 }{ 12 } + \frac{ -17 }{ 3 } + \frac{ 11 }{ 2 } + \frac{ -25 }{ 2 }\)

(ii) \(\frac{ -6 }{ 7 } + \frac{ -5 }{ 6 } + \frac{ -4 }{ 9 } + \frac{ -15 }{ 7 }\)

(iii) \(\frac{ 3 }{ 5 } + \frac{ 7 }{ 3 } + \frac{ 9 }{ 5 } + \frac{ -13 }{ 15 } + \frac{ -7 }{ 3 }\)

(iv) \(\frac{ 4 }{ 13 } + \frac{ -5 }{ 8 } + \frac{ -8 }{ 13 } + \frac{ 9 }{ 13 }\)

(v) \(\frac{ 2 }{ 3 } + \frac{ -4 }{ 5 } + \frac{ 1 }{ 3 } + \frac{ 2 }{ 5 }\)

(vi) \(\frac{ 1 }{ 8 } + \frac{ 5 }{ 12 } + \frac{ 2 }{ 7 } + \frac{ 7 }{ 12 } + \frac{ 9 }( 7 } + \frac{ -5 }{ 16 }\)

 

Solution:

(i) \(\frac{ 11 }{ 12 } + \frac{ -17 }{ 3 } + \frac{ 11 }{ 2 } + \frac{ -25 }{ 2 }\)

= \(\left (\frac{ 11 }{ 2 } + \frac{ -25 }{ 2 } \right ) +  \frac{ 11 }{ 12 } + \frac{ -17 }{ 3 } \)

= \(\frac{ 11 – 25 }{ 2 } + \frac{ 11 }{ 12 } + \frac{ -17 }{ 3 }\)

= \(\frac{ -14 }{ 2 } + \frac{ 11 }{ 12 } + \frac{ -17 }{ 3 }\)

= \(\frac{ -14 \times 6 }{ 2 \times 6 } + \frac{ 11 }{ 12 } + \frac{ -17  \times 4 }{ 3 \times 4 }\)

= \(\frac{ -84 }{ 12 } + \frac{ 11 }{ 12 } + \frac{ -68 }{ 12 }\)

= \(\frac{ -84 + 11 – 68 }{ 12 } \) = \(\frac{ -141 }{ 12 }\) = \(\frac{ -47 }{ 4 }\)

(ii) \(\frac{ -6 }{ 7 } + \frac{ -5 }{ 6 } + \frac{ -4 }{ 9 } + \frac{ -15 }{ 7 }\)

= \(\left (\frac{ -6 }{ 7 } + \frac{ -15 }{ 7 } \right ) +  \frac{ -5 }{ 6 } + \frac{ -4 }{ 9 } \)

= \(\frac{ -6 – 15 }{ 7 } + \frac{ -5 }{ 6 } + \frac{ -4 }{ 9 }\)

= \(\frac{ -21 }{ 7 } + \frac{ -5 }{ 6 } + \frac{ -4 }{ 9 }\)

= \(\frac{ -21 \times 18 }{ 7 \times 18 } + \frac{ -5 \times 21 }{ 6 \times 21 } + \frac{ -4  \times 14 }{ 9 \times 14 }\)

= \(\frac{ -378 }{ 126 } + \frac{ -105 }{ 126 } + \frac{ -56 }{ 126 }\)

= \(\frac{ -378 – 105 – 56 }{ 126 } \) = \(\frac{ -539 }{ 126 }\) = \(\frac{ -77 }{ 18 }\)

(iii) \(\frac{ 3 }{ 5 } + \frac{ 7 }{ 3 } + \frac{ 9 }{ 5 } + \frac{ -13 }{ 15 } + \frac{ -7 }{ 3 }\)

= \(\left (\frac{ 3 }{ 5 } + \frac{ 9 }{ 5 } \right ) + \frac{ 7 }{ 3 } + \frac{ -13 }{ 15 } + \frac{ -7 }{ 3 } \)

= \(\frac{ 3 + 9 }{ 5 } + \frac{ 7 }{ 3 } + \frac{ -13 }{ 15 } + \frac{ -7 }{ 3 }\)

= \(\frac{ 12 }{ 5 } + \frac{ 7 }{ 3 } + \frac{ -13 }{ 15 } +  \frac{ -7 }{ 3 }\)

= \(\frac{ 12 \times 3}{ 5 \times 3 } + \frac{ 7 \times 5 }{ 3 \times 5 } + \frac{ -13 }{ 15 } + \frac{ -7  \times 5 }{ 3 \times 5 }\)

= \(\frac{ 36 }{ 15 } + \frac{ 35 }{ 15 } + \frac{ -13 }{ 15 } + \frac{ -35 }{ 15 }\) = \(\frac{ 36 + 35 – 13 – 35 }{ 15 } \) = \(\frac{ 23 }{ 15 }\)

(iv) \(\frac{ 4 }{ 13 } + \frac{ -5 }{ 8 } + \frac{ -8 }{ 13 } + \frac{ 9 }{ 13 }\)

= \(\left (\frac{ 4 }{ 13 } + \frac{ 9 }{ 13 } + \frac{ -8 }{ 13 } \right ) + \frac{ -5 }{ 13 } \)

= \(\frac{ 4 + 9 – 8 }{ 13 }  + \frac{ -5 }{ 13 }\) = \(\frac{ 5 }{ 13 } + \frac{ -5 }{ 8 } \)

= \(\frac{ 5 \times 8 }{ 13 \times 8 } + \frac{ -5 \times 13 }{ 8 \times 13 } \)

= \(\frac{ 40 }{ 104 } + \frac{ -65 }{ 104 } \) = \(\frac{ 40 – 65 }{ 104 } \) = \(\frac{ -25 }{ 104 }\)

(v) \(\frac{ 2}{ 3 } + \frac{ -4 }{ 5 } + \frac{ 1 }{ 3 } + \frac{ 2 }{ 5 }\)

= \(\left ( \frac{ 2 }{ 3 } + \frac{ 1 }{ 3 }  \right ) + \left( \frac{ 2 }{ 5 } + \frac{ -4 }{ 5 } \right )\)

= \(\frac{ 2 + 1 }{ 3 }  + \frac{ 2 – 4 }{ 5 }\)

= \(\frac{ 3 }{ 3 } + \frac{ -2 }{ 5 } \)

= \(\frac{ 3 \times 5 }{ 3 \times 5 } + \frac{ -2 \times 3 }{ 5 \times 3 } \)

= \(\frac{ 15 }{ 15 } + \frac{ -6 }{ 15 } \)

= \(\frac{ 15 – 6 }{ 15 } \) = \(\frac{ 9 }{ 15 }\) = \(\frac{ 3 }{ 5 }\)

(vi) \(\frac{ 1 }{ 8 } + \frac{ 5 }{ 12 } + \frac{ 2 }{ 7 } + \frac{ 7 }{ 12 } + \frac{ 9 }{ 7 } + \frac{ -5 }{ 16 }\)

= \(\frac { 1 }{ 8 } + \left ( \frac{ 5 }{ 12 } + \frac{ 7 }{ 12 }  \right ) + \left( \frac{ 2 }{ 7 } + \frac{ 9 }{ 7 } \right ) + \frac{ -5 }{ 16 }\)

= \(\frac{ 1 }{ 8 } + \frac{ 5 + 7 }{ 12 }  + \frac{ 2 + 9 }{ 7 } + \frac{ -5 }{ 16 }\)

= \(\frac{ 1 }{ 8 } + \frac{ 12 }{ 12 } + \frac{ 11 }{ 7 } + \frac{ -5 }{ 16 } \)

= \(\frac{ 1 \times 42 }{ 8 \times 42 } + \frac{ 12 \times 28 }{ 12 \times 28 } + \frac{ 11 \times 48 }{ 7 \times 48 } + \frac{ -5 \times 21}{ 16 \times 21 } \)

= \(\frac{ 42 }{ 336} + \frac{ 336 }{ 336 } + \frac{ 528 }{ 336 } + \frac{ -105 }{ 336 } \)

= \(\frac{ 42 + 336 + 528 – 105 }{ 336 } \) = \(\frac{ 801 }{ 336 }\) = \(\frac{ 267 }{ 112 }\)