Question

plz give 100% correct answer do not cheat​

Answer 1

Given:

Value of x = 2/3, y = -5/6 and z = 5/9.

To Verify:

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

Verification:

Put the values in the given equation.

=> x*(y+z) = (x*y)+(x*z)

=> 2/3*[(-5/6)+5/9] = [(2/3)*(-5/6)]+[(2/3)*(5/9)]

Now, take the LCM of 6 and 9.

LCM of 6 and 9 is 18.

=> 2/3*[(-5*3)+(5*2)/18] = [(2*(-5)/3*6]+(2*5/3*9)

=> 2/3*(-15+10/18) = (-10/18)+(10/27)

Now, take the LCM of 18 and 27.

LCM of 18 and 27 is 54.

=> 2/3*(-5/18) = (-10*3+10*2/54)

=> 2*(-5)/3*18 = -30+20/54

=> -10/54 = -10/54

=> -5/27 = -5/27

=> LHS = RHS

Hence verified.

Answer 2

Given :

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

where, x = 2/3, y = -5/6, z = 5/9

Solution :

Let us put the values in the equation to check whether it is equivalent or not :

$\frak{x \times (y + z) = (x \times y) + (x \times z)}$

Putting the values we get

$\frak{ \frac{2}{3} \times ( \frac{ - 5}{6} + \frac{5}{9}) = ( \frac{2}{3} \times \frac{ - 5}{6} ) + ( \frac{2}{3} \times \frac{5}{9} ) }$

For the left hand side the L.C.M is 18

$\frak{ \frac{2}{3} \times ( \frac{ - 5 \times 3}{6 \times 3} + \frac{5 \times 2}{9 \times 2} ) = \frac{ - 5}{9} + \frac{10}{27} }$

Now, taking the L.C.M on the right hand side which is 27

$\frak{ \frac{2}{3} \times ( \frac{ - 15}{18} + \frac{10}{18} ) = \frac{ - 5 \times 3}{9 \times 3} + \frac{10 \times 1}{27 \times 1} }$

$\frak{ \frac{2}{3} \times ( \frac{ - 5}{18} ) = \frac{ - 15}{27} + \frac{10}{27} }$

$\frak{ \frac{ - 5}{27} = \frac{ - 5}{27} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \cal{ \orange{\underbrace{ \red{HENCE, VERIFIED !!}}}}$

Shortcut trick :

• x × (y + z) = (x × y) + (x × z)

• x(y + z) = xy + xz

By factorising the right hand side we get

• x(y + z) = x(y + z)

• LHS = RHS

Hence, Verified ✓✓