Question

verify that |x+y| < |x|+|y| when. x=8/-3 AND Y =-7/9
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Answer 1

Answer:

Given

x=4/9,y=−7/12,z=−2/3

x−(y−z)



=(x−y)−z

L.H.S.=x−(y−z)

=4/9−{−7/12−(−2/3)}

=4/9−(−7/12+2/3)

On further calculation, we get

=4/9−{(−7+8)/12}

=4/9−(1/12)

=4/9−1/12

Taking L.C.M. we get,

=(16−3)/36

=13/36

Now,

R.H.S=(x−y)−z

={4/9−(−7/12)}−(−7/12)

=(4/9+7/12)+7/12

On further calculation, we get

={(16+21)/36}+7/12

=37/36+7/12

Again taking L.C.M. we get,

=(37+21)/36

=58/36

Therefore, x−(y−z)



=(x−y)−z.

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Answer 2

Step-by-step explanation:

< |x|+|

y| wat |x+y| < |x|+|y| when. x=8/-3 AND Y =-7/9

hen. x=8/-3 AND Y =-7/9