# Introduction to Systems of Linear Equations (MUST SEE)

(D4M) — Here is the video transcript for this video.

00:00
in this video you’ll get a must-see introduction to systems of linear equations
but did you know that while people have been solving equations for thousands of
years it wasn’t until descartes invented cartesian geometry that computing
intersections of lines and planes became practical by solving systems of linear
equations so let’s see how hi everyone welcome back in this video
introduction to systems of linear equations is part of the series systems
of linear equations in-depth tutorials for linear algebra
so let’s see what we’re going to cover today
so we’re going to talk about uh two by two systems of linear equations we’re
going to talk about three by three systems of linear equations
and we’re going to talk about m by n systems of linear equations and then
we’re going to see lots of examples all right so let’s go ahead and get started

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so first two by two systems what are two by two systems of linear equations
so um a system of linear equations in two variables we
often use an x and a y when we’re looking at two equations and with two
variables um and then we have the coefficients
here so we have a bunch of uh numbers here um either
for example real numbers or complex numbers
most likely you’ll start off by studying real numbers
now because of the number of letters here used we often use indices here so
how do the indices work so the first indices here the one one
represents the uh first row or the first equation and the first variable
and then we have the one two which is the first
equation in the second variable and then the b1 here is also a number so use the
b’s for the right hand side and then we have the 2 1 which is the

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second equation first variable this is the scalar or the coefficient of x
and the a22 is the coefficient of the y in the second equation and y is the
second variable so this really cuts down on the number
of letters you use we’re just using a’s and b’s with indices that helps a lot
okay for simplicity let’s temporarily assume that the coefficients are
non-zero real numbers and ask the question how many solutions can you get
when you’re looking at a two by two system so for example the key idea is to
realize that each linear equation represents a unique line in the
cartesian plane so if you consider the possible ways
lines can intersect in the plane you come to the conclusion that there must
be no solutions one unique solution or infinitely many points so we can come to
this conclusion by looking at the way lines interact with each other

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in the plane so you may have two parallel lines in which case you would
say there’s no solutions there’s no point on both of these lines
or you may you get one unique solution right there’s the point of intersection
or you may get infinitely many points where the lines are actually right on
top of each other so it may look like you have two separate lines and
algebraically looking you know it may look like you have two different lines
when you actually go graph those lines you actually see that geometrically
they’re the same lines sitting right on top of each other so you can say
instantly many points and we’ll see how to work out some systems that look like
that all right so let’s look at our first
example this is a two by two system how do we solve it so um
i’m going to multiply the first equation by minus 2
and that will give me a minus 4x and then a minus 2 here
so when we’re doing this it’s important to remem to remember to multiply the

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whole equation by a non-zero number so if you go graph this equation and you
get a line and you go graph this equation and you get a line and they’re
the same line geometrically speaking they represent the same line now here
i’m going to use the same equation to represent that line
now you may ask why did i multiply the first equation by -2
so the reason why is because now the x’s match and now when i add them together
i’m going to cancel the x’s or the x’s add up to zero
so here we’re going to get minus y and then we add up the right-hand side
and we get zero so in other words y equals zero
and if y is zero we can go back to any one of these or any one of these and we
can find the x’s so here let’s go to the
top one here so two x plus three times y we just found y is zero
so i’m using the top equation there with y equals zero and now we can solve for

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the x so 2x equals zero in other words x equals zero so this is an example where
we have two lines and they’re intersecting at one point which is zero zero
so we just have two lines intersecting here at zero zero
that’s the unique solution to this system right here
but not all systems have a unique solution sometimes we’ll have infinitely
many solutions sometimes we’ll have no solution sometimes we’ll have exactly
one solution so now let’s consider the three by three
systems of linear equations so what happens when we have a three by
three so three by three system um so now we’ll have obviously three
equations and we have three variables and so you know we’re going to need
these indices to really work out here really nicely
so on the right hand side we have the b’s b1 b2 b3
and those are scalars they’re real numbers

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and then we have the variables x y and z these are the variables and these right
here the a i j are the coefficients so a 1 1 is the
coefficient of the first uh first equation the first variable and this
will be the first equation second variable and this will be the first
equation third variable and then so on so for example this one right here is
the coefficient of the in the third equation of the second variable right there
so this would be a way to write out a three by three system of linear equations
so geometrically these equations under the assumptions that you don’t have a
trivial case like zero zero zeros your scalars right so you get three planes or
you get planes in three dimensions so how many different types of solutions if
you’re if you’re considering planes how many different types of solution sets
could there be so by considering um you know planes intersecting in three

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dimensions you can come to the conclusion that there’s either no
solutions one unique solution or infinitely many points
that solves all three equations right here so there there’s three
possibilities again all right so let’s look at solving this
system right here this is a system with three equations and three unknowns three
variables now one way to solve the system is to
write out a paragraph like like right here i’m going to show you a different way
here in a minute but this is this is one way this is writing out all the words
explaining all the steps and this is the way students usually typically don’t
write don’t like to do this this is solving a system right so we’re going to
say multiply the first equation by -2 and add it to the second equation
obtaining this here multiplying the first equation by m and
so on and you get this paragraph so this is what a solution would look like in
paragraph form but actually what i’m going to do is

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just solve it just by looking at the equations only
and by saying most of the words that’s how most people will
solve these because there’s just too many words and you and it’s just
not as fun as just working out the equation so let’s do this just working
out the equations right here so i’m going to use the strategy though
that we used uh and when i wrote up the paragraph form years i’m going to try to
eliminate the z here so to eliminate the z here
what we’re going to do is we’re going to multiply this first equation right here
yeah so we have this first equation i’m going to multiply this by um
you know let’s try to follow this right here by -2
and then add it to the second equation so we can eliminate the z’s right here

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so let’s do that let’s multiply this equation by minus 2 right here and then
i’m going to um you know eliminate the z right here so
let’s do that here so we’re going to get here minus 2x minus 4y
and then minus 6z equals 0. so that’s multiplying the first equation by
minus two and then i’m going to add it to the
second equation here so i’m just going to keep this as a 4x plus 5y plus six z
equals three and i’m going to keep this equation right here the same
so it’s very important that you don’t that you do operations here that don’t
change the solution set so you know look at this system and look at this system
equations two and three are exactly the same so all i did was multiply the first

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equation by minus two everything both sides by minus two
so these two systems have the same solution set i haven’t changed the
solution set but when i add these two together here
what are we going to get we’re going to get a 2x and we’re going to get a y
and the z’s are going to cancel that’s why i multiplied this one by minus two
so i can add it here so i can cancel the z’s and then we get equals three okay
so now i’m going to multiply the first equation by
minus three and add it to the third equation so i’m gonna multiply by
uh minus three here and we’re going to get this system here now so multiply by
minus 3 and add so i’m going to multiply
this one by minus 3. so i’m going to get minus 3x minus 6y minus 9z equals 0

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and i’m going to keep the rest of the system exactly the same so here we have
the a system with the same solution set but so i just multiplied the top
equation by -3 and the rest are the same equation two and
three are exactly the same now when i add equations one and three together
what are we gonna get so we’re gonna get four x we’re going to get 2y
we’re going to get 0 z’s and then we’re going to get 0 equals 0.
now if i divide this one by 2 everywhere we get 2x plus y equals zero
so here we have a problem so these two right here tell me that
whatever x and y satisfy this system two x plus three y has to be three

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and two x plus y has to be zero and that cannot happen so this
system right here has no solution so there’s an example so if if you can
derive a contradiction then there’s no solution if you try to
find x y and z well whatever x and y you come up with 2x plus y will be 3 and
then 2x plus y will be 0 and that just it doesn’t make any sense so in fact
there can be no xyz in fact there could not be any xy so there cannot be any xyz
okay so now let’s look at what happened if i were to try to solve
this using geometry i mean what if this plane here was graphed in this plane and
this plane what would that look like would you be able to solve it using
kind of geometry sometimes you’re asked to solve two by two systems using
geometry so using geometry would be very difficult here you would have these
three planes here and you would have to rotate and orient it in such a way

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for you to determine if there was a solution or not
moreover when you’re looking at higher ordered systems like 4×4 or 5×6 or 7×8
it’s going to be very difficult using geometry so the point is is that we’re
using algebra to solve geometric questions to answer geometric
questions and the question is do these planes intersect at a unique
point and we just saw using algebra that they do not okay so
let’s do another one here um so this time the a b and c are going
to be hidden from us so they could be anything a could be any
real number b can be any real number c can be any real number now they’re not
variables they’re just hidden numbers okay so what i’m going to do is i’m
going to eliminate the x first so we’re going to um eliminate the x so
i’m going to subtract these first two together what

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happens if you subtract these first two you’re going to get x minus x
and now we’re going to get a minus y so let’s write that over here so minus y
and i’m going to get minus 5z and i’m going to get a minus b
so a minus b is a number we don’t know what that number is because we don’t
know a and b but it’s a number so again i’m just subtracting x minus x is 0.
2y minus 3y and 3z minus 8z and then on the right hand side a minus b
now we can subtract the first and the third
so that’ll give us x minus x which is zero
2y minus 2y that will give us zero we’re going to get 3z minus 2z which is 1z
and we’re going to get a minus c so we found z
now we can use this one right here to go
back and plug in this z here and then to find the y

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so let’s do that here so minus y equals let’s move that 5z to the other side so
5z and then plus a and then a minus b and then let’s solve for y
multiply everything both sides by negative minus 5 times z minus a plus b
now let’s come in with the z here so y will be minus five times the z
minus a plus b so now we’re going to get minus 5a plus 5c minus a plus b
so i’ll distribute the minus 5. and now we’re going to get the minus 6a and
plus a b plus a 5c so there we go so there’s our y there
there’s our z there’s our y now we can go to any one of these here
and find the x so which one looks easiest um

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how about the third one so the third one says that x is equal to c
minus move the 2y over minus 2y and move the 2z over
and now substitute in the y this is the y right here so minus 6 a
plus b plus five c and then now minus two times z
so minus two and then a minus c that will be the x there
so what do we get for this x here so we’re going to get c plus 12a minus 2b

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minus 10c minus 2a plus 2c so that’s a minus 10c how many a’s do we get
we’re going to get 10 a’s and how many b’s are we going to get minus two b’s
and how many c’s we’re going to get minus so minus nine minus seven c’s
okay so very good so we’ll write the solution like this x is
10 a minus two b minus seven c and then the y is uh
now i think i’ll just put the boxes around them here’s the x here’s the y
and here’s the z so convey coming back to the original
problem now if someone gives you a 1 1 1 you’ll just come down here and plug in
the 1 1 1 you’ll have the x plug in the 1 1 1 you’ll have the y and plug in the

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1 and the 1 and you’ll have the z plug in any a b and c you want
we’ll find the x we’ll find the y we’ll find the z in other words
this system right here has exactly one solution
if you tell me what the a b and c is i’ll be able to tell you what that one
unique solution is so that’s a really fun problem there i
really like that problem okay so now let’s talk about m by n systems
so m by n systems here and now we’re going to talk about um
we know we now we need n variables so x1 x2 all the way to xn and
they’re going to you know be variables and we’re going to need some
coefficients here so the number so xj is in the ith equation so the i it
comes by the equation and j is the variable and that and so the a i j is the

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coefficient of the jth variable and the ith equation so that’s exactly the way
we’ve been using them before so we’ve got some real numbers here b
for the right hand side of our equations so we have m equations so we have n
unknowns and m equations and so a system will look like this
so we have the first equation right here in variables x1 x2 xn
and we have the coefficient so notice each coefficient starts with one each
index here starts with one one one first variable second variable nth
variable and b1 and then we have the exact same thing
but now we have the second equation so second these all start with twos
and so this is b2 and then you continue all the way down to the nth

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so this is a system of simultaneous linear equations or sometimes you’ll
just say systems of linear equations or even in this video or as long as you
understand the context you could just call it a system okay so here’s a
m by n linear system a solution of this system is an ordered
set of numbers right so the first um first one would be you know something
like this like x1 x2 so instead of xy you’d have x1 x2 and
then the last one would be xn and so whenever you put things in order
like this this is the first one representing the first variable there and
each of these each a solution would have to satisfy all m
equations or all m statements so you’d have to be able to plug the solution in
into the first one and you get out b1 plug this into the second equation and

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you get out b2 and so on so it’s called consistent
if there is at least one solution and sometimes systems will have infinitely
many solutions but if there are no solutions at all
it’s called an inconsistent uh system okay so let’s look at this one right
here solve this linear system right here this is a two by three
so we have two equations in three variables
okay so to solve this system right here um we notice that the z’s here
so that’s what we’re going to do here we’re going to add these up and cancel
out the z’s and see what happens so if we add them up here we’re going to get
minus 100x and if we add these up here we’re going to get plus 600y
we add the z’s they add up to zero and then the right hand side adds up to two

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hundred so here um you know we can simplify this by a hundred
and then just get minus x plus six y equals two
and so that’s a nice equation there now which one should we let be free
right this has infinitely many points on it and
infinitely many x and y satisfy this equation so
which one should we be let be free well we can let x be free or we can that y be
free so um i’m gonna let the y be free because if i move the y over
then i won’t have to divide by anything in front of x and so i’ll get cleaner
solutions but i’ll actually do it both ways so you can see what i see what i’m
saying here so i’m going to say let x be free here so let x equal t be a real
number or you might say that differently you might just say

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let y equal to t be free so let x be a real number or you might
even say let x be equal to t be an arbitrary real number
any any one it’s fine but now what you get for the other one if x is free what
will the y be so if x is free then i’ll move x over and i’ll solve for y
so six y will be equal to two plus move the x over
and then the x is a t so plus t and so y here will be two plus t over six
over here if y is t then i’ll move the y over and i’ll say minus x equals two
and when i move the six y over it becomes minus six
and then the y is a t so minus six t and so now i’m going to multiply through
by a negative and i’m going to say 6t minus 2.
so i like this one better because it doesn’t have any fractions in it

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but now that i know the x and now that i know the y
whichever route you want to take you can go back and find the z
in fact you can use either one to find a z so now i can split this off and solve
for z on either one but either way you would get the same answer for z
so with this y and this x and you plug it into either one you’ll get the same
so i’m going to use this one right here so z is equal to what
so z is equal to 200 and then move the minus 100y
to the other side and y is t in this in this case and then minus 50x
so minus 50. uh sorry but i’m moving this over yeah
it’s moving it over it’s minus 50 and then the x is this uh
six t minus two so six t minus two so we get two hundred

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and then we’re going to get here um minus 100 t
and what do we get here minus 300 t and then plus 100
so in this case we’re going to get 300 and then minus 400 t
and so we could write the solutions like this we have the x
which is six t minus two we have the y’s of t
and we have the z which is three hundred minus four hundred t
and remember t is a real number so you choose any real number you want
you plug it in here here and here and you’ll have a solution you’ll have an x
y z that will solve both of these equations right here
now if you wanted to continue with this route
you would get in a solution set also you would get all the solutions

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but this time your t would be over here would be a t and this would be your y
and then you could use either one of these to go find the z and that would
give you a z now your solution set will look
different but it will actually be the same x y and z
it’ll be the same solution set so remember this is any t any any real
number you can plug in here so there’s infinitely many now when you when you
come up with an infinite set like this and using a parameter
you you get infinitely many solutions to the system right here
okay good i hope that helps and now let’s look at some more examples here
so um we’re going to go a little bit quicker now
so we’ve got this system right here and we’re going to try to solve it
and i’m looking at the z’s right here and looking at the x’s right here and
i’m thinking like what is the best strategy here so what i’m going to do is

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i’m going to multiply this top equation by minus 3 and this equation by -3 and
try to eliminate some of the x’s here and so here’s what we’re going to get
we’re going to get minus 3x minus 6y minus 9z equals minus 117. so that’s by
multiplying by 3 everything here now i’m going to multiply by um
sorry by by minus three now i’m going to multiply the second one also by minus
three and i’m going to get minus three x and minus nine y and minus six 6z
and minus 102. and so i multiplied by minus 3 minus 3 minus 3 and minus 3.
now i’m going to keep the third one the same now it’s important to remember that
your solution set doesn’t change so here’s another three by three system
but they have the same solution set how do i know that because all i did was

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multiply the first equation by minus three that doesn’t change that plane
whatever that plane is you can go graph it multiplying it by the equation by
minus three doesn’t change the geometric object
i’ll multiply this one by minus three also okay but now we can add them
together we’re going to get minus 4y minus 8z
equals minus 91. and for the second one we get minus 7y minus 5z
equals -76 right so adding them together here and here
we’re going to get the minus 4y minus 8z and here we get minus 91.
and now when i subtract here we’re going to cancel out the x’s and we get minus
7y minus 5z minus 76. so we went from a 3×3 system to a 2×2 system
two by two systems are easier so now i’m going to try to get a 28 here

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in front of the y’s here so i’m going to get a 28y plus 56z equals 637.
so can you tell what i did there right multiply by
uh minus seven minus seven and minus seven
now to get a 28 here i’m going to with the negative 28
i’m going to get a minus 28 so because i’m multiplying by minus four
or sorry a four so i’m going to get a minus 20 z
and my minus 76 and then i’m going to get a 304 minus 304
so this is this is a two by two system but these have the same solution sets
here is all i did was multiply by constant i multiplied this plane by constant
right so now the y’s will cancel and now we can add these up together so we’re

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going to get 36 z equals 333 and if you divide you’re going to get 9.25
9.25 there divide 333 by 36 now we can come back and find the y
or sorry find the yeah find the y we already found the z so now we’re going
to find the y so i’m going to look right here and say minus 7y
and then i’m going to say equals 2 minus 76 and i’m going to move the 5 z
over so 5z but the z here is 9.25 so y is minus 76 plus 5 9.25 all divided by -7
so using a quick calculator on that we’re going to get 4.25
4.25 okay so we found the z we found the y now we can come back to any of these

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here and find the x so let’s use the second one here so x equals 34 minus 3y
minus 2z right moving moving all that stuff over minus 3y minus 2z
now plugging in what we know for the y and what we know for the z
so x equals 34 minus 3 times y which is the 4.25 minus 2 times z 9.25
and if we calculate that out using calculator we’re going to get 2.75
so we found the one and only one solution to this so that’s 2.75 4.25
and then the 9.25 there’s the one and only one solution the solution
all right very good now in the next couple videos we’re

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going to get a lot better at that because
what is the strategy going on right here it seems so haphazard
so there’s a way to formalize the strategy and make it easier to follow
and make it easier so that you can put it on a calculator or or calculator or
computer and in the computer can do the work for you
gaussian elimination will be coming up and we’ll see how to do all this but
first we’re going to learn how to put this in matrix form and
this so you won’t have so many x’s y’s floating around and all this work so
let’s look at one more example real quick what happens if this um system changes
right here now we have zero three and six here
and so what can we do with this here so i’m going to um
look at these z’s right here and i’m going to try to make them all match so

00:33
this is just one strategy here um so i’m gonna multiply this top one by
here by a minus three and this one by a minus three and this one by a
two so i’m going to multiply this by 2. so i’ll give you a minus 18 minus 18 and
that’ll give me an 18. so what will happen if i do all that
let’s check it out so the first one i’m going to get here is minus 6x minus 12y
minus right so we’re multiplying by minus 3 minus 18z equals 0.
and then on the second one we get minus 12x minus 15y and then minus 18z
and then we’re going to get minus nine and on the third one i’m going to
multiply by two so 14x plus 16y and then plus 18 z and then equals to 12.

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so there’s our new system it has the exact same solution set as the original
okay and so now i’m going to work on these z’s here and eliminate them
so when i add the uh first and the third i’m going to get eight x plus four y
equals under uh sorry twelve right and then um adding the um second and the
third we’re going to get the 2x plus the y equals the 7. okay
uh sorry three that’s not a seven all right so now what i’m gonna do is
i’m going to divide this by four here and i’m going to get 2x plus y equals
three but i already have a 2x plus y equals 3.
so now the 2×2 system really is just one equation that means there’s going to be
infinitely many solutions i need to choose one of them to be free

00:35
because i don’t want fractions i’m going to let the the x be free so here we go
let x be equal to t be a real number all right so then what will the y be
the y will be 3 minus the 2x but the x is a t
right so that’s why i chose the x to be free because then the y looks really
nice now let’s go up in here and find the z here so we can use any one we want
so let’s use the top one here so what are we going to get well we’re
going to get the 6z is equal to 0 minus the 2x minus the 4y and using the
x that we have so this will be minus 2 times the x which is a t

00:36
minus 4 times the y so six z is what just minus two t minus twelve plus eight t
so z is um minus two t oh sorry six t minus 12 over six
so in other words t minus two so z is t minus two
so we have infinitely many solutions t and then the y is three minus two t
and then the z is t minus two and then t is any real number here
any real number at all all right so that was fun and let’s move on to the

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last example here now so now what i’m going to try to do is try to eliminate the
z’s again here the z’s look like very nice instead of 1
4 7 i’m going to look at 3 6 and 9. so i’m going to multiply this one here by my
minus 3 and add it to this one right here
so i’m going to multiply by minus 3 i’m going to get minus 3x minus 6y
and then minus 9z equals 0. and then i’m going to keep multiply this one here by
we can do a minus two and add it to the so i’ll just keep this one the same here
four x plus five y plus six z equals three now what happens when we add these
together so we get x so let’s see here when i multiply by

00:38
minus 3 okay so i’m adding to 7x plus 8y plus 9z equals 0.
these are what i’m adding together so i get 4x
and then plus 2y and then the z’s cancel out and i get 0.
or chopping it by 2 we get 2x plus y equals zero
okay so that comes from the first and the third
now what happens if we do the first and the second so now i’m gonna multiply
this by minus two so i’m gonna get minus two x minus four y
minus six z equals zero and i’m gonna keep the second one the same
and now what happens when we add them together
so we get two x and here we get a plus y and z and equals to 3.
and so right here we can see that these two right here say there’s no solution

00:39
in other words the system is inconsistent there we go
okay so if you have any questions or ideas i hope that you use the comment
section below and like this video if you got value out of it
don’t forget this series uh this video is part of a series systems of linear
equations in-depth tutorials for linear algebra and i want to say thank you for
watching and i hope to see you next time bye-bye
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