Question

I verify the distributive property a x (b+c) a (axb)+(axc) for the rational numbers
(1) a= -1/2 b=2/3 and c= -5/6
(2)a=4/7 b=1/5 and c=3/4​

Answer 1

Given:

Distributive property a×(b+c)=(a×b)+(a×c)

(1) a= -1/2   b=2/3   c=-5/6

To find:

We have to verify the distributive property for the given a,b,c values

Solution:

The given distributive property is a×(b+c) = (a×b)+(a×c)

Let us substitute the given values in above equation

$(\frac{-1}{2})$ × $(\frac{2}{3}+ \frac{-5}{6} )$              = $(\frac{-1}{2}$ × $\frac{2}{3} )$ + $(\frac{-1}{2}$ × $\frac{-5}{6} )$

$(\frac{-1}{2})$ × $(\frac{(2)(2) + (1)(-5)}{6} )$     =  $(\frac{-2}{6} )$ + $(\frac{5}{12} )$

          $(\frac{-1}{2})$ × $(\frac{4-5}{6} )$        =  $\frac{(2)(-2)+(1)(5)}{12}$

             $(\frac{-1}{2})$ ×$(\frac{-1}{6} )$        =   $\frac{-4+5}{12}$

                          $\frac{1}{12}$        =    $\frac{1}{12}$    

                     L.H.S       =    R.H.S

Hence verified.

(2) a=4/7   b=1/5   c=3/4

To find:

We have to verify the distributive property for the given a,b,c values

Solution:

The given distributive property is a×(b+c) = (a×b)+(a×c)

Let us substitute the given values in above equation

       $(\frac{4}{7})$ × $(\frac{1}{5} + \frac{3}{4})$     =  $(\frac{4}{7}$ ×$\frac{1}{5} )$ + $(\frac{4}{7}$ ×$\frac{3}{4} )$

$(\frac{4}{7})$ ×$(\frac{(4)(1)+(5)(3)}{20} )$     =  $(\frac{4}{35} )$ + $(\frac{12}{28} )$

           $(\frac{4}{7})$ × $(\frac{19}{20} )$      =  $(\frac{(4)(4)+(5)(12)}{140})$

                  $(\frac{76}{140} )$      =  $(\frac{76}{140} )$  

                 L.H.S     =   R.H.S

Hence verified.

Answer 2

Step-by-step explanation:

Distributive property a×(b+c)=(a×b)+(a×c)

(1) a= -1/2   b=2/3   c=-5/6

To find:

We have to verify the distributive property for the given a,b,c values

Solution:

The given distributive property is a×(b+c) = (a×b)+(a×c)

Let us substitute the given values in above equation

(\frac{-1}{2})(2−1) × (\frac{2}{3}+ \frac{-5}{6} )(32+6−5)              = (\frac{-1}{2}(2−1 × \frac{2}{3} )32) + (\frac{-1}{2}(2−1 × \frac{-5}{6} )6−5)

(\frac{-1}{2})(2−1) × (\frac{(2)(2) + (1)(-5)}{6} )(6(2)(2)+(1)(−5))     =  (\frac{-2}{6} )(6−2) + (\frac{5}{12} )(125)

          (\frac{-1}{2})(2−1) × (\frac{4-5}{6} )(64−5)        =  \frac{(2)(-2)+(1)(5)}{12}12(2)(−2)+(1)(5)

             (\frac{-1}{2})(2−1) ×(\frac{-1}{6} )(6−1)        =   \frac{-4+5}{12}12−4+5

                          \frac{1}{12}121        =    \frac{1}{12}121    

                     L.H.S       =    R.H.S

Hence verified.

(2) a=4/7   b=1/5   c=3/4

To find:

We have to verify the distributive property for the given a,b,c values

Solution:

The given distributive property is a×(b+c) = (a×b)+(a×c)

Let us substitute the given values in above equation

       (\frac{4}{7})(74) × (\frac{1}{5} + \frac{3}{4})(51+43)     =  (\frac{4}{7}(74 ×\frac{1}{5} )51) + (\frac{4}{7}(74 ×\frac{3}{4} )43)

(\frac{4}{7})(74) ×(\frac{(4)(1)+(5)(3)}{20} )(20(4)(1)+(5)(3))     =  (\frac{4}{35} )(354) + (\frac{12}{28} )(2812)

           (\frac{4}{7})(74) × (\frac{19}{20} )(2019)      =  (\frac{(4)(4)+(5)(12)}{140})(140(4)(4)+(5)(12))

                  (\frac{76}{140} )(14076)      =  (\frac{76}{140} )(14076)  

                 L.H.S     =   R.H.S

Hence verified.