Question

Verify:x+(y+z)=(x+y)+zbytakingx=−3/4,y=−3/8 and z=14/15

Answer 1

Given :- Verify:x+(y+z)=(x+y)+z by taking x = −3/4, y = −3/8 and z = 14/15 .

Solution :-

→ x + (y + z) = (x + y) + z

putting given values,

→ (-3/4) + (-3/8 + 14/15) = (-3/4 - 3/8) + (14/15)

→ (-3/4) + (-3*15 + 14*8)/120 = (-3*2 - 3)/8 + (14/15)

→ (-3/4) + (67/120) = (-9/8) + (14/15)

→ (-3*30 + 67)/120 = (-9*15 + 14*8)/120

→ (-23/120) = (-23/120) (verified)

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Answer 2

Given:

Verify: x+(y+z)=(x+y)+z

By taking x=−3/4, y=−3/8 and z=14/15

Solution:

x+(y+z)=(x+y)+z

LHS = x+(y+z)

Substituting the given values.

$= > \dfrac{ - 3}{4} + ( \dfrac{ - 3}{8} + \dfrac{14}{15} )$

$= > \dfrac{ - 3}{4} - \dfrac{ 3}{8} + \dfrac{14}{15}$

LCM is 120

$= > \dfrac{ - 90 - 45 + 112}{120}$

$= > \dfrac{ - 23}{120}$

_______________________________

RHS: (x+y)+z

Substituting the given values.

$= > ( \dfrac{ - 3}{4} +( \dfrac{ - 3}{8} )) + \dfrac{14}{15}$

$= > \dfrac{ - 3}{4} - \dfrac{ 3}{8} + \dfrac{14}{15}$

LCM is 120

$= > \dfrac{ - 90 - 45 + 112}{120}$

$= > \dfrac{ - 23}{120}$

Therefore, LHS = RHS.

Hence, Verified.