Given:
[tex]a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}[/tex]
To verify:
[tex]a(b-c)=ab-ac[/tex] for the given values.
Solution:
We have,
[tex]a=1.6,b=\dfrac{1}{-2},c=\dfrac{-5}{-7}[/tex]
We need to verify [tex]a(b-c)=ab-ac[/tex].
Taking left hand side, we get
[tex]a(b-c)=1.6\left(\dfrac{1}{-2}-\dfrac{-5}{-7}\right)[/tex]
[tex]a(b-c)=1.6\left(-\dfrac{1}{2}-\dfrac{5}{7}\right)[/tex]
Taking LCM, we get
[tex]a(b-c)=1.6\left(\dfrac{-7-10}{14}\right)[/tex]
[tex]a(b-c)=\dfrac{16}{10}\left(\dfrac{-17}{14}\right)[/tex]
[tex]a(b-c)=\dfrac{8}{5}\left(\dfrac{-17}{14}\right)[/tex]
[tex]a(b-c)=-\dfrac{68}{35}\right)[/tex]
Taking right hand side, we get
[tex]ab-ac=1.6\times \dfrac{1}{-2}-1.6\times \dfrac{-5}{-7}[/tex]
[tex]ab-ac=-\dfrac{16}{20}-\dfrac{8}{7}[/tex]
[tex]ab-ac=-\dfrac{4}{5}-\dfrac{8}{7}[/tex]
Taking LCM, we get
[tex]ab-ac=\dfrac{-28-40}{35}[/tex]
[tex]ab-ac=\dfrac{-68}{35}[/tex]
Now,
[tex]LHS=RHS[/tex]
Hence proved.
