Reducing Linear Equations into Simpler Form Video Lecture | Advance Learner Course: Mathematics (Maths) Class 7

good morning children now we know how to
solve a linear equation but you know if
the equation is not in linear form we
can also reduce a given equation to
linear form and then solve it so let us
see here we have equation Z plus 15
divided by Z is equal to 3 by 4 now is
this a linear equation
no why if you see the expression on left
hand side that is Z plus 15 by Z is not
a linear equation but you know we can
reduce this equation to the linear form
let us see now let's try to reduce it to
the linear form z plus 15 by z is equal
to 3 by 4 first we multiply both the
sides by Z you may ask why we need to do
that this is because if we multiply put
the elitists and the RHS by Z then we'll
this can be simplified as Z in the
denominator would go off so let's do it
Z multiplied by Z plus 15 divided by Z
is equal to said x 3 by 4 let's simplify
this so desert said gets cancelled and
what are we left with we are left with Z
plus 15 is equal to 3 said by 4 or you
can say 3 by 4 z it's the same thing
next we do the cross multiplication to
simplify this and what do we do then
this here we can say the denominator is
1 and we'll do the cross multiplication
so here the LHS would be multiplied by 4
which was the divisor here and 1 which
is the divisor here
would become the multiplicand here so we
would get on the RHS 3z x 1 so 4 x z
plus 15 is equal to 3 z x 1 if you
notice we have put brackets here which
means 4 needs to be multiplied by both Z
and also 15 so let's do it we get 4 Z
plus 4 multiplied by 15 becomes 60 equal
to 3z next if you see this equation is
in linear form why because Z has only
the power which is equal to 1 now let's
solve it because we already now know how
to solve a linear equation here we have
4 Z plus 60 is equal to 3z again what we
should do subtract 3 set from both sides
why we are just following that to
simplify we should bring the variable on
one side so if we subtract 3 set from
both the sides what do we get for Z plus
60 minus 3 Z is equal to 3 Z minus 3 Z 4
Z minus 3 Z becomes Z and plus 60 we
bring it down and on the RHS what do you
see 3 said minus 3 Z equal to 0 next
what do we do we subtract 60 from both
the sides so that Z is there on the left
hand side alone with no other constant
on the side so Z equal to minus 60 how Z
plus 60 - 60 becomes Z as 60 - 60 cancel
each other and when we subtract 60 from
0
get zero minus 60 which is equal to
minus 60 on the RHS so here we see we
get the solution which is Z equal to
minus 60 again we will check if we got
the right answer or not so putting the
value of Z in the given equation the
equation is Z plus 15 by Z is equal to 3
by 4
let's put Z equal to minus 60 so putting
the value what do we get Z replaced by
minus 16 plus 15 bring it down
divided by Z so divided by minus 60 is
equal to 3 by 4 solving it what do we
get minus 60 plus 15 becomes minus 45
and minus 60 asses on the LHS on the RHS
we already have 3 by 4
simplifying this further - - gets
cancelled and 45 by 60 can also be
reduced so here we get 15 threes are 45
and 15 for a 60 so you see 3 by 4 is
equal to 3 by 4 so LHS equal to RHS and
we got the solution so now you know how
to reduce a given equation to the linear
form that's all for now bye bye children