Question

VERIFY the statements, by taking x=-1/3,y=-3/5,z=-4/9 1.x×(y+z) =x×y+x×z

Answer 1

Given Values ⇒

x = -1/3,
y = -3/5
z = -4/9

Given Equations :-
x(y + z) = xy +xz

Substituting the given values in the L.H.S. Side of the Equation,

x(y + z) = -1/3(-3/5 + (-4/9)
= -1/3(-3/5 -4/9)
= -1/3{(-27 - 20)/45} 
= -1/3(-47/45)
= 47/135

Now, Solving the R.H.S. Side of the equation,
xy + xz = -1/3(-3/5) + (-1/3)(-4/9)
= 1/5 + 4/27
= (27 + 20)/135
= 47/135

∵ L.H.S. = R.H.S.
∴ Given Equation is 100% correct.

Hence, It is verified.


Hope it helps.

Answer 2

x(y + z) = xy + xz , this algebraic operation is known as $\underline{{\text{distributive law}}}$ .

Given, x = -1/3 , y = -3/5 and z = -4/9
LHS = x(y + z)
= (-1/3)[-3/5 + ( -4/9) ]
= (-1/3) [ -3/5 - 4/9]
= (1/3) [ 3/5 + 4/9]
= (1/3) [(3 × 9 + 4 × 5)/5 × 9]
= (1/3) [ 47/45 ]
= 47/(3 × 45 )
= 47/135

RHS = xy + xz
= (-1/3) × (-3/5) + (-1/3) × (-4/9)
= 3/15 + 4/27
= 3 × 9/135 + 4 × 5/135
= (3 × 9 + 4 × 5)/135
= 47/135
∵ LHS = RHS
Hence, it is verified that x(y + z) = xy + xz