Solving Equations Having the Variable on both Sides Video Lecture | Advance Learner Course: Mathematics (Maths) Class 7

hello friends this video on linear
equations in one variable part 4 is
brought to you by exam feel calm no more
fear from exam so let us now try to
solve linear equations with variables on
both sides because till now it was quite
simpler when you have constants on one
side now if you have variables on both
sides then how do you go about it so let
us take an example of this type of an
equation which has variables on both
sides so let's consider this 5x plus 3
is equal to X plus 10 so you have
variable here you have variables here so
how do we handle this so again here also
we will follow the same trick we will
try to collect all the variables on the
left hand side we will try to send all
the constants to the right hand side so
for that what we would do
5x minus X so this X would have come
this side and this 3 would go the side
so this would become 10 minus 3 so 5x
minus 1x would be 4 X 10 minus 3 would
be 7 so therefore X will be equal to 7
by 4 so this would be a solution for
this equation so you see approach
remains the same it is just that you
need to remember those simple tips when
you transpose constants and variables
from one side to another so here what we
are going to do is we will arrange all
those with variables on one side and we
keep all the constants on the other side
so that is the simple mantra that we
will follow to solve linear equations so
let us look at some other examples so
let's say you have an equation 5t minus
3 is equal to 3t minus 5 so here also we
will say the variable here we will
saying the constant here so we get 5t
minus 3 t is equal to minus 5 plus 3 or
we can say 2 t is equal to minus 2 or we
can say t is equal to minus 2 divided by
2 or this is equal to minus 1 therefore
t is equal to minus 1 is a solution for
this equation so keeping these points in
mind
try to solve some more equations with
variables on both sides question number
one so on 4z plus 3 is equal to 6 plus
2z so in this case we collect all the
variables on one side and the constants
on the other so we get 2 z is equal to 3
or we can say Z is equal to 3 by 2 so
this is the solution for this equation
question number 2 solve X is equal to 4
by 5 into X plus 10 so in this case what
are we going to do so what we're
building is to to make it simpler
because due to the presence of 4 by 5
physic region is like looking a little
different so what we will do is we will
send this 5 on the side so that we do
not have this complexity of having a
denominator extra so what happens this
becomes 5x is equal to 4x plus 10 so now
we will follow the same approach so let
us first find out the variable and the
constant terms separately like here we
have done 4 X and plus 40 so now let's
bring the variables on one side and the
constant of the other so we get X is
equal to 40 question number 3 so on 2 X
by 3 plus 1 is equal to 7 X by 15 plus 3
here we gain the same thing let's bring
the terms with X on left hand side so it
is 2 X by 3 minus 7 X by 15 is equal to
3 minus 1 so here 3 and 15 the LCM
wouldn't 15 so this would be 10 X minus
7 X and this would be equal to 2 so this
is 3 X divided by 15 is equal to 2 all
we can say 3 X would be equal to 2 into
15 or 3 X is equal to 13 therefore we
can say X is equal to 30 divided by 3 so
3 10 to 30 so X is equal to
ten would be a solution for the
situation question number four so on to
y plus 5x3
is equal to twenty six by three minus y
so here again 2y plus y is equal to 26
by three minus 5 by 3 so this becomes 3
y is equal to 21 by 3 or Y is equal to
here 3 7 to 21 so this would be 7 and
this 3 will come here it'll become 7 by
3 so y equal to 7 by 3 would be a
solution for the situation question
about 5 3 M is equal to 5 M minus 8 by 5
so here again we bring all the MS on
left hand side so this is what we get
not 3 M minus 5 m would be minus 2 M is
equal to minus 8 by 5 or M would be
equal to minus 8 by 5 into 1 by minus 2
so 2 fours are 8 this minus and minus
will also get cancelled therefore M is
equal to 4 by 5 as a solution for this
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