As the property states (x + y) + z = x + (y + z)
Use the values as such,
\( \dfrac{-2}{5} + \dfrac{4}{3} + \dfrac{-7}{10} =\dfrac{ -2}{5 }+ \dfrac{4}{3} + \dfrac{-7}{10}\)
Let us consider LHS \(( \dfrac{-2}{5} + \dfrac{4}{3} )+ (\dfrac{-7}{10})\)
Taking LCM for 5 and 3 is 15
\(\dfrac{ (-2× 3)}{(5×3)} + \dfrac{(4×5)}{(3×5)}\)
\( \dfrac{-6}{15} + \dfrac{20}{15}\)
Since the denominators are same we can add them directly,
\( \dfrac{-6}{15} + \dfrac{20}{15}= \dfrac{(-6+20)}{15} = \dfrac{14}{15}\)
\( \dfrac{14}{15}+\dfrac{-7}{10}\)
Taking LCM for 15 and 10 is 30
\(\dfrac{ (14×2)}{(15×2)} + \dfrac{(-7×3)}{(10×3)}\)
\(\dfrac{ 28}{30 }+ \dfrac{(-21)}{30}\)
Since the denominators are same we can add them directly,
\( \dfrac{(28+(-21)}{30} = \dfrac{(28-21)}{30} = \dfrac{7}{30}\)
Let us consider RHS \(\dfrac{ -2}{5} + \dfrac{4}{3} + \dfrac{-7}{10}\)
Taking LCM for 3 and 10 is 30
\( \dfrac{4}{3} + \dfrac{-7}{10} = \dfrac{(4×10)}{(3×10)} + \dfrac{(-7×3)}{(10×3)}\)
= \(\dfrac{ 40}{30 }+ \dfrac{(-21)}{30}\)
Since the denominators are same we can add them directly,
\( \dfrac{ 40}{30 }+ \dfrac{(-21)}{30}= \dfrac{(40-21)}{30} = \dfrac{19}{30}\)
\( \dfrac{-2}{5} + \dfrac{19}{30}\)
Taking LCM for 5 and 30 is 30
\( \dfrac{-2}{5} + \dfrac{19}{30}=\dfrac{(-2×6)}{(5×6) }+\dfrac{ (19×1)}{(30×1)}\)
=\(\dfrac{ -12}{30} + \dfrac{19}{30}\)
Since the denominators are same we can add them directly,
= \(\dfrac{ (-12 + 19)}{30} = \dfrac{7}{30}\)
LHS = RHS associativity of addition of rational numbers is verified.
Answered by Sakshi | 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 \(\dfrac{5 }{ 2} and \dfrac{– 11 }{ 12}\)