Solving Simple Algebraic Equations - Simple Equations Video Lecture - Grade

okay so welcome back and in this video
we are going to use the concept of
balance and observation to solve some
simple equations so let us start with
the first equation which says that X
plus 3 equals 8 now here our left hand
side is X plus 3 and the right hand side
is 8 so we know that to solve this
equation we have to isolate the X and
how do we isolate this X is depends on
this term here so we have left hand side
which is X plus 3 now if I want to just
get X i can simply subtract three from
this equation right so subtracting
three from
LHS what we get is x plus 3 minus 3 so
these two cancel out and what we get is
simply X now we have the right hand side
as well the RHS and RHS is 8 so in this
case also because we have to maintain
the balance we need to subtract the
right hand side also three from the
right-hand side so we get 5 therefore
the new equation which we get is x
equals 5 now this is your solution of
this simple equation and how do i check
whether the solution is correct or not
so to check the solution we substitute
substitute x equals 5 in the original
equation
so in this case it would be X plus 3
equals 8 and putting wherever we are X
we just put 5 so this becomes 5 plus 3
equals 8 or when this evaluates it
becomes 8 equals 8 which means that the
left-hand side equals right hand side
this is the solution now I just want you
to reflect a bit what we have done here
is we just isolated the eggs and found
out a value of x which is like five and
then when we substitute this we see that
putting that value of x gives us the
same value that we get on the other side
of the equation which is the right hand
side so this is what we mean by equation
being a condition on a variable right X
simply cannot have any value right it
cannot have any value for this equation
to be satisfied the equation here let's
say that if i take x equals 1 okay if i
take x equals 1 so in that case my left
hand side is going to be equal to 1 that
is X plus 3 which is one plus three
equals four but if you see the right
hand side it is equal to eight now for
is not equal to eight right this means
that this equation doesn't work this is
not really an equation it's X is equal
to one is not a condition that satisfies
the equation x equals 5 is the condition
that it satisfies this equation and
therefore it is called the solution of
the equation so this is another way to
look at what we mean by the solution of
an equation let us look at next
example and in this case we will try to
use addition operation to solve an
equation so we have X minus 3 equals 10
and as usual i would like to isolate the
value I elect isolate X so we have X
minus 3 equals 10 and this seems pretty
straightforward that if we subtract or
we add sorry if we add 3 to both our HS
that is the right hand side and the left
hand side LHS so let's start with the
left hand side which is X minus 3 now
let us add 3 to the sides of X minus 3
plus 3 that gives us X because this 3
will get cancelled out then we have the
right hand side which is 10 and now we
are supposed to add a value of 3 so we
have 10 plus 3 equals 30 okay so now we
have the new equation which says that X
is equal to 13 and this is our solution
now let us try to check the solution
that whether X equals 13 is the correct
solution or not so what we do is simply
substitute x equals 13 in the equation X
minus 3 equals 10 so this is our left
hand side and this is the right hand
side of this equation now this X here
becomes 30-3 this right hand side still
equal to 10 30-3 is 10 equals 10
therefore you can see that the left hand
side equals to the right hand side
therefore this solution x equals 13 is
correct and x equals 13 satisfy this
equation X minus 3 equals 10 so this is
an example where we use addition and
observation to find out the solution to
a simple equation now let us look at
another example and in this example we
will try to use the division method so
in this case we have 5 x equals 35 now
note that in earlier examples are used
to do it separately that is LHS I used
to do it separately and Rho separately
that was mainly for the purposes of
explanation now I will be a bit more
concise so i can write this is LHS and
this is our HS okay and then the
equation is 5y equals 35 now I want to
find out a solution to why a value of y
which actually satisfies this equation
now for that what I need to do is to
isolate y on the side and if you look at
it closely if i divide the left hand
side by five if we divide this by 55
gets cancelled and the numerator and the
denominator and we have y but when we
divide a lectures by five we also need
to divide the right hand side by five
right because we need to maintain the
balance of the equation
therefore let us try to divide left-hand
side and right hand side by funny to do
that so we have 5 x divided by 5 equals
35 / 5 or this gets cancelled and we
have y equals 5 times 7 is 35 so we have
y equals 7 now this is what we got as a
solution to our equation but is that
correct so for that we just check the
solution and we check the solution by
substituting y equals 7 in the equation
original equation so we have LHS and we
have archers so we have 5y equals 35 and
when I put seven heroes over here 5 x 7
equals this is still 35 and when I
evaluate this 35 equals 35 therefore our
solution y equals 7 is correct and the
value of y being 7 satisfies this
equation so in this example we used just
division to solve the simple equation of
5y equals 35 now let us look at the last
operation which is remaining which is
that of using multiplication so we have
an equation where in says that M
x 2 equals Phi and you want to find out
a value of M which satisfies this
equation so let us write down this as
the left-hand side and the right hand
side now if you observe this equation
how can isolate them that is what we
have been doing all along so the way to
isolate m would be to multiply the
left-hand side by two now if we multiply
the left-hand side by two i need to
multiply the right hand side by two as
well to maintain the balance so let us
do that so let me just write down that
multiplying our HS and LHS by five so we
have n by 2 sorry it's not five it's
going to be too it's too because we need
so multiplying both of them by two so n
by 2 into 2 and 5 into 2 right so this 2
gets cancelled and we have M equals ten
so this is our new equation now this is
the solution to the equation mm equals
10 so let us check by substituting M
equals ten in the equation so we have
LHS we have a right right hand side so
we have em by two equals five so let's
put 10 M equals 10 so 10 x 2 equals 5 or
we get let me just write it over here 5
equals 5 this is my LHS
and this is my RHS so you see that the
left hand side is equal to right hand
side and therefore M equals ten is a
solution to the equation so in this
video we have just picked up very simple
equations which involved just one basic
operation you just had to observe it and
there was just one basic operation
either addition multiplication division
or subtraction to isolate X and find out
the solution and we also learned how to
check the solution in the next video we
will take up next step of examples
wherein we might have to use more than
one operation to isolate X and come to a
solution