Hello! Mr. Tarrou. Today we are going to talk about solving linear systems of equations.
If you solve a linear system of equations, you are looking for where two lines cross.
There are three situations there. You can have two lines that never cross because they
are parallel. That will be called no solution or inconsistent which is a term our new textbooks
use. But no solution when they are parallel. They do not cross. You can have a situation
where you have a line and then the other line lands right on top of it. That is going to
be called Many Solutions, or as our books call it Dependent. And then of course as this
one will, you can simply have a situation where yes the two lines are going to cross
at a certain (x,y) value. Which we are going to find two ways with this one example. Now
these which have one solution are kind of like normal and things kind of happen like
you would expect. But when you have two lines which are parallel, you are going to be solving
your problems and all of a sudden everything just goes away. Like all of the x's and y's
are going to just disappear and you might have an answer like 0 equals 4. So all of
a sudden it seems like you have done something wrong and what is left is a false statement.
That is what happens when your lines never cross. There is no solution so you end up
with a statement which is false. That is the algebra going no... no... those two lines
are parallel. They never cross each other. If your two lines are lying right on top of
each other, called many solutions, or dependent... and I really prefer no solution, one solution,
or many solutions. When you are solving these questions, and you won't see this example
today, but you are doing your algebra work and all of the x's are going to go poof and
disappear. Well when they do that, if you get a true statement like 0 equals 0, or 5=5,
then that means that it does not matter what x you plug in or what points you look at along
the lines... those two lines are sharing all of the same points because they are the exactly
the same line once they are graphed. So that Dependent or Many Solutions. And if you are
doing your algebra and all of the x's and y's just sort of disappear, then that is your
case for many solutions. Let's find out where these two lines intersect. Now you have two,
or three actually, methods of solving systems of equations. You can graph the lines through
y=mx+b and just visually see where those two lines cross. That is not very accurate if
the point is a difficult one, especially if there are decimals. So we have Substitution
method and Linear Combination method. I am going to show you the substitution method
first. Now the reason why this is a good fit for substitution method is because I have
some terms whose coefficients are one. Now I have two choices and it does not matter
how I start. With substitution your first step is to get one equation set equal to one
variable. So i am going to use the first one, just because I felt like it. So 3x-y equals
3. I am going to get this equation solved for the term with the coefficient of 1. I
am going to subtract both sides by 3x. And then get rid of the -1 in front of the y by
dividing everything by -1. So y equals 3x-3. Great! That does not really tell me an answer
because that is only the first step of doing substitution. Get an equation set equal to
one of its variables. Step two is to take this expression that you have and here it
is either going to be equal to y or x, depending on whatever problem you are doing. Now plug
this into the other equation, the one that you have not touched yet. So that is going
to be the second one in this case. It is going to be x, this is step two... the actual substitution
process, x plus two times y... or 2 times something. The y is the place where you plug
stuff in. So I have taken the y out and put in a set of parenthesis. In its place is 3x-3.
I am going to solve this for the remaining variable which in this case is x. That is
going to involve getting the distributive property started. x+6x-6 equals 8. Combine
like terms if you have them. 7x-6 equals 8. I am going to add 6 to both sides and get
7x equals... Maybe I should show that. Add 6 to both sides. And 7x equals 14. You divide
by 7 and x is equal to 2. And so this is my second step of doing the Substitution Method.
Get one equation solved for one of the variables. Plug that into the other equation that you
have not worked with yet and find half of your solution. Find the x... in this case
it was the x of the coordinate where the two lines cross. Well, then how do you find the
y? Because an intersection point is an x and a y value. Well it is easy. Take this answer
and plug it back into one of your original equations. So the third step of substitution
is to take this answer and plug back into one of the original ones. I am going to pick
the one with all positive numbers. x plus 2y equals 8. I am going to plug in 2 for x.
So 2+2y equals 8. I am going to subtract both sides by 2 and get 2y=6. Divide by 2 and I
get y=3. So these two lines cross at the point of (2,3). BAM! And if you saw a picture that
is where those two lines would be crossing. Now, let's see what that would like through
the Linear Combination process. Ok, now that was the Substitution Process. Go back and
replay it and write down the steps that I verbally stated for the process of substitution.
For the Substitution Method you want to take one of the equations and get it equal to one
of the variables. Solve for a variable that has a coefficient of 1, or just a coefficient
of 1. Take your expression you found and plug it into the other equation, solve for the
remaining variable, and then plug your answer into one of the original two equations. It
does not matter which of the two you pick. Substitution method is very very powerful
and sometimes it is the best choice of solving a system of equations. However, if none of
your coefficients are equal to one and they are all values... like this could be 4 and
5, and this could be 2 and this is already 2. If all of your coefficients of x and y
are... well... if none of them are one then trying to solve by substitution method will
cause you to use fractions or decimals through the entire process which can make things complicated.
So let's talk about the Linear Combination method. To do Linear Combination, what you
want to do is simply add those equations and do it in such a way that one of the variables
will cancel away. So to a Linear Combination, what you want to do is look at your two equations
and decide what do I want to get rid of. Well, we are going to get rid of the y's first.
Once you decide.. with your Linear Combination method... choose what variable you want to
get rid of first. Then your goal is to think of... ok... I want to multiply one or both
of my equations by a number so that I can create a pair of opposite numbers. I want
to do the same problem so you see I get the same answer again. So I have a negative one
and a positive two in front of my y variables. I have just arbitrarily decided that is what
I want to get rid of first. I need to do some kind of multiplication to create a pair of
opposite numbers so that when I add I get zero. Well I already have two. And this negative
one... -1's are easy to change. If I have a positive two and I want this to be an opposite,
or negative two... and I am looking for the lowest common multiple, then what can I multiply
to -1 to get -2 the opposite of positive two? We are going to multiply that by positive
two. And I am going to multiply this through the entire equation, going all the way through.
So we are going to get 2 times 3 which is 6x. 2 times -1 giving -2y. And 2 times 3 which
is 6. So everything in there got multiplied by 2. Now I going to just copy the second
equation. x+2y=8. See the opposite numbers that are in front of the y that I arbitrarily
decided I wanted to get rid of first? Once you get that opposite pairs of numbers in
front of one the two variables, then you can add those equations and get 6x and 1x is 7x,
6 and 8 is 14. Oh! Boom! Just like that. So this is going to be a bit less work than the
example I just erased. 7x=14. I am going to solve for x by undoing that multiplication
by division. And x equals 2. The same answer as it was a minute ago. Ok. Linear Combination
one more time. Pick what variable you want to get rid of. Do some kind of multiplication
to get that opposite pair of coefficients that you want. Once you get those opposite
numbers you add the equations and then solve for the remaining variable. That last step
should be pretty quick. Ok, so this is one half of the point where the lines intersect.
I find the other half by once again plugging in the x value, or y if that is what you found
first, into one of the two original problems. And let's see. Let's pick the first one just
to see if it matters which one of these equations we use. 3 times 2 minus y equals 3. 3 times
2 is 6, now minus y equals 3. Subtract both sides by 6. Negative y equals -3. Negative
y means you have -1 times y. You want to undo that multiplication of -1. You do that with
division, and y equals 3. Guess what? That is exactly the same answer that we had a second
ago. These two lines cross at the point of (2,3). Ok, I have three minutes left. Let
me do one more example before I get out of here. When you talk about linear equations,
you have all of these x's and y's floating around. That is all fine and dandy, but then
you get something that almost seems too easy. Like 3x... Not too easy but it is like there
is stuff missing, which makes them confusing. So here is one equation and here is the other
one. Solve this system of equations. Now linear combination though I do prefer that method
of solving systems of equations would be horrible for this one. Well... Ok.. Maybe not horrible,
but I think it would be more confusing for students to try to figure this out. Sometimes
we have ways that are good for teaching and makes it less confusing, and then we have
our ways to do these problems that don't work well for teaching and get across what you
are doing. So... How do I take and do this problem through the Substitution Method? Well
the Substitution Method is first get an equation solved for one of the variables. haha. I have
already got that. That is why substitution is good for this one. Second step, take that
expression where the y is by itself... whatever variable you have... In this case it is y
where it is by itself. Take its value and plug it into the other equation. So the substitution
step here is... I am going to take the y which is 5 and plug it into the other equation.
So 3x-5 equals 4. That is 3x, after I add both sides by 5 which some of you might need
to see... And I should be doing this as a teacher. And I want to undo the multiplication
of 3. I am going to do that with division. And where this line and this line cross the
x value is going to be 3. Well what is the y value? Right there. It told me. Y is equal
to 5. So the point where these two lines cross is (3,5). And I don't think I have time to
draw this in the video, but this is just a normal line with some kind of slope. What
kind of equation.. Or excuse me. What kind of line do you get from y=5? What do those
graphs look like? Don't forget y equals a number is a horizontal line. This horizontal
line is going to go through the y value of 5. And horizontal lines have a slope that
is equal to zero. If that where x=5, x equals a number is a vertical line. They have an
undefined slope because it is rise over the run value of zero, and you cannot divide by
zero. So y equals a number is a horizontal line, x equals a number is a vertical line,
and if you have x and y's both in your equation well then you just have a normal line line
y=mx+b line slope and y intercept. Alright. So, solving system of equations. If that was
enough examples, leave a comment and I will do another one and I will do some more video
examples for you. I hope this helped. WHOOO!!! Have a great day:D