Question

Verify (y+z)=xy+xz, where x=1 upon ,y=1 upon 3, z=1 upon 4
plz answer it is very urgrnt

Answer 1

Answer:

xy + x + y = 23\brSpace xy+x+y+1=24 \brSpace (x+1)(y+1)=24..........................1\brSpace xz + x + z = 41\brSpace xz + x + z + 1 = 42\brSpace (x+1)(z+1) = 42..............2\brSpace yz + y + z +1 = 28\brSpace (y+1)(z+1) = 28...................3\brSpace from 1, 2 & 3 ony x=5,y=3 and z=6 satisfy the given equations therefore\brSpace x + y + z=14

Answer 2

Answer:

LHS = RHS = 7/12

y+z = xy + xz = 7/12

Step-by-step explanation:

Given,

x = 1

y = 1/3

z = 1/4

To prove : y + z = xy + xz

Proof :

LHS = y + z = 1/3 + 1/4

On taking the LCM of 3 and 4, we get

1/3 + 1/4 = 4/12 + 3/12 = 7/12

Hence,

LHS = 7/12 ..... (i)

Now,

RHS = xy + xz = (1×1/3) + (1×1/4)

= 1/3 + 1/4 = 4/12 + 3/12 = 7/12

Hence

RHS = 7/12 .... (ii)

From equation (i) and (ii), we get

LHS = RHS = 7/12

Hence,

y+z = xy + xz = 7/12

Hence, verified....... :)