Firstly group the rational numbers with same denominators
\( \dfrac{11}{12 }+ \dfrac{-17}{3} + \dfrac{11}{2} +\dfrac{ -25}{2} \)
\(\dfrac{ 11}{12} + \dfrac{-17}{3} +\dfrac{ (11-25)}{2}\)
\(\dfrac{ 11}{12} + \dfrac{-17}{3} + \dfrac{-14}{2}\)
By taking LCM for 12, 3 and 2 we get, 12
\(\dfrac{ (11×1)}{(12×1) }+\dfrac{ (-17×4)}{(3×4) }+ \dfrac{(-14×6)}{(2×6)}\)
\(\dfrac{ 11}{12 }+ \dfrac{-68}{12} +\dfrac{ -84}{12}\)
Since the denominators are same can be added directly
\(\dfrac{ (11-68-84}{12} = \dfrac{-141}{12}\)
Answered by Aaryan | 1 year agoBy what number should 1365 be divided to get 31 as quotient and 32 as remainder?
Which of the following statement is true / false?
(i) \(\dfrac{ 2 }{ 3} – \dfrac{4 }{ 5}\) is not a rational number.
(ii) \( \dfrac{ -5 }{ 7}\) is the additive inverse of \( \dfrac{ 5 }{ 7}\)
(iii) 0 is the additive inverse of its own.
(iv) Commutative property holds for subtraction of rational numbers.
(v) Associative property does not hold for subtraction of rational numbers.
(vi) 0 is the identity element for subtraction of rational numbers.
If x = \( \dfrac{4 }{ 9}\), y =\( \dfrac{-7 }{ 12}\) and z = \( \dfrac{-2 }{ 3}\), then verify that x – (y – z) ≠ (x – y) – z
If x = \(\dfrac{ – 4 }{ 7}\) and y = \( \dfrac{2 }{ 5}\), then verify that x – y ≠ y – x
Subtract the sum of \(\dfrac{ – 5 }{ 7} and\dfrac{ – 8 }{ 3}\) from the sum of \(\dfrac{5 }{ 2} and \dfrac{– 11 }{ 12}\)