Graphical Solution of Linear Inequalities in 2 variables Video Lecture | Mathematics (Maths) for JEE Main & Advanced

hi there now in this video what I want
to do is show you how you shade a region
given by an linear inequality of two
variables of the form ax plus B y is
greater than or less than a given
constant and there's two examples I've
got here one where it's zero and the
other where it isn't zero and I'm going
to show you how we handle each of these
different types and I also encourage you
to try these two examples here which
just revise the methods shown in these
two examples now in this first example
we're asked to shade the region given by
this linear inequality here 2x minus y
is greater than or equal to zero now
whenever you've got zero on the end what
I would suggest you do is make Y the
subject
so therefore what we've got is 2x is
greater than or equal to Y if we add Y
to both sides and then if I bring Y to
the left hand side we end up with Y
being less than or equal to 2x now when
we have it in this format the next step
is to draw the graph of y equals 2x so
it's a straight line graph gradient is
two and it passes through the origin now
in order to find the region that
satisfies this inequality or satisfies
any of these because they're equivalent
to one another you just take the point
any point on either this side of the
line or on this side of the line it's
totally up to you I'll demonstrate by
showing you a couple of points now
rather than take a point somewhere on
the grid here it's a lot easier to take
a point on the coordinate axes
so I'm going to take this point on the
coordinate axis which is clearly on this
side of the line let's say I call that
point a and it's coordinates are going
to be 0 1 x is 0 Y is 1 and if I take
that point what I do is I test it out
I'll show you would just put here test
the point a which has the coordinates
then naught 1 and I can test this point
either in this inequality up here which
I'll call 1 or let's say I could test it
in this inequality here which I'll call
2 so let's say we start by substituting
x equals naught and y equals 1 in the
inequality 1 and if that's the case then
if X is not this term is 0 and then we
get minus 1 and then we have minus 1 is
greater than or equal to 0 which is a
false statement so I just write false
there so what that's telling me then is
that any points on this side of the line
are going to give false results if we
put the x and y values into here the
region that we want that gives true
results is going to be this region over
here you'd get exactly the same result
if you substituted X is naught and y is
1 into this inequality 2 I'll show you
so if we were to substitute our results
then into equation 2 then we have 4 y
that's going to be 1 and we've got less
than or equal to 2 times the x value 2 x
naught which is not + 1 clearly isn't
less than or equal to 0 so it's false
resulting in the fact that the gain
results what we need to shave is in here
but you might not have chosen say the
point a on the y-axis let's suppose you
took another point say on the x-axis
let's say you took this point here we'll
call this point B and it's coordinates
then would be 1 0 and if that were the
case you'd be testing the point B so
let's say we test the point B with
coordinates 1 0 so X is 1 Y is 0 and
let's say we substitute those values
then in the inequality one and doing
that we end up with 2 times 1 which is 2
minus the Y value which is 0 so we've
got 2 is greater than or equal to 0 so 2
is greater than or equal to 0 and as an
inequality that is true and if you had
decided to substitute it in the
inequality 2 then it doesn't make any
difference because here you'd have for Y
naught is less than or equal to 2 times
1 naught is less than or equal to 2
which is a true statement so if you had
taken the point B to test then this is
in a true region and it would result in
shading this area here as you can see so
several ways then that you could
establish this shading by taking points
such as a on the y-axis or point B on
the x-axis and if you just take say a
point any point in this region here
let's suppose you took this point here
with coordinates 2 1 if you were to
substitute that into here 2 2 so 4 minus
1 is 3 which is greater than or equal to
0 so you can see it works and it would
be true for all points in this region
now in this next example got to shade 2x
plus 3y is less than 6 and with this
example it's not got a zero on the end
like in this one so what we do is one of
two things you can either draw the graph
from this stage or you can make Y the
subject if you make Y the subject let's
just put this down it's going to be the
same as Y as being less than minus 2/3 X
and then plus 2 but when it comes to
drawing this graph y equals minus 2/3 X
plus 2 it's not so easy as doing this
method which I'll show you and that is
to say that when x equals 0 work out
what Y will be when x is 0 you get 3 y
would equal 6 and so did I number 3 Y
would equal 2 and then we set y equal to
0
so when y equals 0 we end up with 2x
equaling 6 divide both sides by 2 and
you get x equals 3 so if you draw the
axes and put the point when X is not Y
is 2 on not 2 it's going to be that
point there just put a 2 in and when y
is not X is 3 that's going to be that
point there then the line is going to
look something like this and notice I've
drawn it there's a dotted one as opposed
to a solid one as we had here and that's
because the inequality that we're using
is a less than the same would be true if
it was just greater than it's always
solid though if you've got an equals
underneath okay whether it's greater
than or equal to or less than or equal
to draw a solid line now I could have
drawn the line from this particular
inequality it's got the form
y equals MX plus C it would across the
y-axis that to which you can see it does
the gradient would be minus two-thirds
which is let's say a little bit more
awkward to do not impossible for every
three units across you drop two units so
you could work from that version but
when it doesn't equal zero on the end I
prefer this particular method of drawing
the line now when it comes to testing
which region which side of this line we
have to shade it's very easy when you've
got a line that doesn't pass through the
origin as we had in this example because
the point that you should test is the
origin itself so let's say we test the
origin Oh chess coordinates naught
naught X is not wise not obviously and
we substitute this into the inequality
up here which I'll call one so sub in
one or you could substitute it into this
second inequality it's up to you which
I'll call two so if your I sub in one
let's put sub in two at the same time
here okay but you're only gonna want to
choose one of these versions then if you
substitute not not into the one here
you've got naught plus naught which is
naught is less than 6 so naught is less
than six which is a true statement if
you had gone for two then you've got
here naught is less than naught plus two
and that with naught is less than two
which again as expected should be the
same answer it's true so that means that
the origin or any point on this side of
the line is going to satisfy this
inequality so what we shade is this okay
now I've got two more examples here that
I would encourage you to try so do pause
the video when you come back you can
check your work solutions against mine
okay welcome back then
had a go now with this one shade 5 y -2
X is greater than or equal to 10 because
it doesn't equal zero then when it comes
to sketching the graph I'm going to
adopt this method here it's a lot easier
so if we draw the axes on a grid then
for the point where this crosses the x
axis set y equal to zero when y equals
zero if we have minus 2x equals 10 then
the X would equal minus 5 so we can mark
that point on here at minus 5 and then
for where it crosses the y axis set X
equal to zero so 5 y would equal 10 y
would equal 2 so if we go up to 2 there
and Mark that on so drawing a line
through those two points gives us this
5y minus 2x equals 10 then now because
it doesn't pass through the origin then
we can test using that point so test
when x is 0 and y 0 so if I substitute
that into there I end up with getting 0
minus 0 is greater than or equal to 10
so 0 is greater than or equal to 10
which is clearly a full statement so
that means that the region that we want
is not on this side of the line but over
here now in the next example we've got
shade y plus x is less than 0 and
because this has a 0 on the end trying
to set x is 0 Y is 0 is pointless
because we know it goes through the
origin so I'd want to rearrange this to
make Y the subject and so Y would be
less than minus X so need to draw the
graph of y equals minus X and that's
going to be a graph looking like this
notice I've drawn a dotted line here
because we've got a less
than now because it passes through the
origin I need to take a point on one of
the sides of this line preferably on an
axis as I said earlier it makes it
easier it's up to you which point you
take but I'm going to take this point
here which I'll call a and that has
coordinates naught 1 so I'm testing that
point let's say we're testing the point
a with coordinates naught 1 and so if I
substitute that result into here then
I've got for Y which is 1 is less than
minus X that would be 1 is less than 0
which is clearly false so that means
this point a is not in the shaded region
the shaded region then must be here ok
well I hope that sets you up now for
shading linear inequalities then in two
variables
you