Examples: Properties of Rational Numbers (Part 1) Video Lecture - Class 8

hello friends this video on rational
numbers 515 is brought to you by exam
for calm normal fear for exam on the
properties of rational numbers which we
have studied so far let us try to solve
a few questions using appropriate
properties find out the solution for
minus 2 by 3 multiplied by 3 by 5 plus 5
by 2 minus 3 by 5 multiplied by 1 by 6
so how can we make use of the property
so let us first find that if we have
something in common between two terms so
when you compare this term with this
term we see 3 by 5 is common so we will
just change the order a bit so minus 2
by 3 into 3 by 5 and this term that is
we will put it as plus and then we will
keep it as 3 by 5 into minus 1 by 6 so
this will make it all the more easier so
this was minus sign so if you consider
it as a plus so basically the minus sign
goes either to the minus 3 or to the
minus 6 so I have transferred to 2 minus
6 to keep it more simple plus 5 by 2 so
this term so this is what we get
now this was done by commutative
property because in commutative property
we know that a plus B is always equal to
B plus a for rational numbers so by
commutative property with this we did
this so now we will take help of the
distributive property now from
distributive property we know that a in
B plus a into C is equal to a into B
plus C so we will make use of this
property where three by five is our a so
this will become 3 by 5 into B plus C
that is minus 2 by 3 minus 1 by 6 plus 5
by 2 so this will be equal to 3 by 5
into minus 5 by 6 plus 5 by 2 so here 5
+ 5 will cancel out 3 2 2 6 so this is
minus 1 by 2 plus 5 by 2 so this is
equal to 4 by 2 2 3 so 4 so this will be
2 so this would be our answer
so in most of the cases variable you
need to make use of the properties to
solve an expression first of all try to
see in which of the terms you have
something in common because the moment
you are able to find out something in
common you will be very easily able to
apply the distributive property and
distributive property will make your
calculations a lot easier let's look at
question number 2 here again using the
properties we will have to solve this so
let's try to solve it so it is 2 by 5
into minus 3 by 7 minus 1 by 6 into 3 by
2 plus 1 by 14 into 2 by 5 so here again
you have three terms this is one term
this is another term and this is the
third though where do you see something
in common so 2 by 5 is a common term
between these the first term and the
last term so let's change the order so
by commutative property we can say that
2 by 5 multiplied by minus 3 by 7 plus 1
by 14 into 2 by 5 minus 1 by 6 into 3 by
2 now even when we change the order the
values will remain the same that is
assured by the commutative property of
addition so now in these two terms we
will try to apply the distributive
property so in these two terms how do we
apply distributive property the common
term is 2 by 5 which is our a so a is B
plus C so B is minus 3 by 7 and C is 1
by 14 so this is minus 1 by 6 into 3 by
2 so here 3 2 so 6 so here we get 2 by
five into this one piece minus six plus
one divided by 14 minus 1 by 4 so this
would be equal to 2 by 5 into minus 5 5
45 and 5 will cancel out two seconds of
working this man this one by 4 so this
would be equal to minus 1 by 7 minus 1
by 4 so when you take their insulin you
get minus 4 1 7 divided by 28 at this
value comes out to be minus 11 by 28 so
this is the solution so he will be made
use of the distributive property here we
made use of the commutative property so
by making use of these properties we can
simply find our solutions it thank you
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