Question 2 iiiFind-6-13-7-15

6/13 ( 7/15 )= 6 ×15/13×15 ( 7× 13/15 × 13 )= 90/195 ( 91/195 ) [L.C.M of 13 and 15 is 195] = 90 ( 91)/195 = 90 + 91/195 = 1/ ...

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Byju's Answer

Standard VI

Mathematics

Addition of Fractions

Question 2 ii...

Question

Question 2 (iii)

Find:
$\frac{- 6}{13} - (\frac{- 7}{15})$

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Solution

$\frac{- 6}{13} - (\frac{- 7}{15}) = \frac{- 6 \times 15}{13 \times 15} - (\frac{- 7 \times 13}{15 \times 13}) = \frac{- 90}{195} - (\frac{- 91}{195})$ [L.C.M of 13 and 15 is 195]
$= \frac{- 90 - (- 91)}{195} = \frac{- 90 + 91}{195} = \frac{1}{195}$

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