Question

verify the following ​

Answer 1

Answer: True (L.H.S. = R.H.S.)

Step-by-step explanation:

Given,

L.H.S,

$\displaystyle=(\frac{-7}{11}+\frac{2}{-5})+\frac{-13}{22}\\\;\\=(-\frac{7}{11}-\frac{2}{5})-\frac{13}{22}\\\;\\=-(\frac{7}{11}+\frac{2}{5})-\frac{13}{22}\\\;\\=-(\frac{7\times5+11\times2}{55})-\frac{13}{22}\\\;\\=-(\frac{35+22}{55})-\frac{13}{22}\\\;\\=-(\frac{57}{55})-\frac{13}{22}\\\;\\=-\frac{57}{55}-\frac{13}{22}\\\;\\=-(\frac{57}{55}+\frac{13}{22})\\\;\\=-(\frac{114+65}{110})\\\;\\=-\frac{179}{110}$

Now,

R.H.S,

$\displaystyle=\frac{-7}{11}+(\frac{2}{-5}+\frac{-13}{22})\\\;\\=-\frac{7}{11}+(-\frac{2}{5}-\frac{13}{22})\\\;\\=-\frac{7}{11}-(\frac{2}{5}+\frac{13}{22})\\\;\\=-\frac{7}{11}-(\frac{22\times2+13\times5}{110})\\\;\\=-\frac{7}{11}-(\frac{44+65}{110})\\\;\\=-\frac{7}{11}-(\frac{109}{110})\\\;\\=-\frac{7}{11}-\frac{109}{110}\\\;\\=-(\frac{7}{11}+\frac{109}{110})\\\;\\=-(\frac{70+109}{110})\\\;\\=-\frac{179}{110}$

Hence, We can see in both cases

L.H.S. = R.H.S.

Answer 2

Solution :-

Given,

$\sf L.H.S = \bigg( \dfrac{- 7}{11} + \dfrac{2}{- 5} \bigg) + \dfrac{-13}{22} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = \bigg( - \frac{7}{11} - \frac{2}{5} \bigg) - \frac{13}{22} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = \bigg( \frac{ - 35 - 22}{55} \bigg) - \frac{13}{22} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = - \frac{57}{55} - \frac{13}{22} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: = - \dfrac{179}{110}$

Now,

$\sf R.H.S = \dfrac{- 7}{11} + \bigg(\dfrac{2}{- 5} + \dfrac{-13}{22} \bigg) \\ \: \: \: \: \: \: \: \: \: \: \: \: \sf = \dfrac{- 7}{11} + \bigg( - \dfrac{2}{5} - \dfrac{13}{22} \bigg) \\ \: \: \: \: \: \: \: \: \: \: \: \: \sf = \dfrac{- 7}{11} - \bigg( \dfrac{2}{5} + \dfrac{13}{22} \bigg) \\ \: \: \: \: \: \: \: \: \: \: \: \: \sf = \dfrac{- 7}{11} - \frac{109}{110} \\ \: \: \: \: \: \: \: \: \: \: \: \: \sf = \dfrac{ - 179}{110}$

Therefore,

L.H.S = R.H.S

Hence Proved !!!!