I know if you go on to MATH110 at least you're gonna have to be able to find
points of intersection of two graphs whether they're lines
or parabolas or whatever. So
Let's get right to a problem
and then we'll review kinda what we did. I've got two equations here.
I know that and these ones happen to be lines.
They're going to intersect in exactly one point in this case. Now,
other types of graphs may intersect in more than one point but in this case we're
looking for one point and remember that one point
has two values an x and a y.
So, we'll start off by solving one of the two equations.
For one of its variables, you get to pick whichever one you want, whichever is
easiest.
Now in this case I think the
blue x is the easiest cuz it's already solved for x.
Okay, so I'm gonna choose the blue x
and then I'm gonna do step b and that is to substitute
the expression equivalent to that variable
3y
into the other equation. I'm substituting it instead of x.
When I say the other equation I hope you know
I'm talking about the green one. So...
let's substitute 3y
instead of x into that green equation we'll get
an equation
with only one variable, only y, which we can solve for.
3y plus y is four y.
If 4y equals 20,
and I divide both sides by four
I know y is 5/
Remember, I'm looking for a point of intersection
and a point has an x and a y so I'm kind of halfway there
as I found the Y but I still have to find the x.
So, what up, but I have more information now.
So what I'll do is take the value of y, which is five,
excuse me, take the value
y, which is five, and substitute it
into the original equation instead of y.
So instead of y, I'm gonna write
5 and now I have an equation with only one variable and it's the
other variable, x, which I can solve.
and x will be...
x will be 15.
So the solution to this system, if you would,
of more than one line is
the point where x equals 15 and y is five or sometimes we write
the point (15,5).
Not so bad. We did it without looking at the graph which is pretty cool.
Since I went to all that work
and I know this works in the blue equation.
I just solved it. It would probably behoove me
to check it in the other equation, the green one. Doesn't take much work, does it?
Hmmm, let's put fifteen in for x
and five in for y and just make sure
And I know it works. In fact I know this is the point where
they intersect cuz this
pair works in both the equations and that's the only point
that works in both equations the point where they intersect.
What happened? Well, let's go over what we did
and pay attention.
First thing we did was we solved one of the two equations
for one of its variables, whichever was easiest. Then we substituted the
expression
equivalent to that variable into the other equation
and that created an equation with only one variable
that we could solve. And we did that; we solved the now single variable equation
created by step b.
That gave us more information and told us what one of the variables was, what's its
value.
So we substituted that value that we found in step C.
into either of the original two equations and solved
for the other variable and then don't forget!
Check your proposed solution. Check
your x and your y into both equations. Make sure you didn't make a
mistake. If it doesn't work in both equations
you haven't found the point of intersection, okay?
This time let's do it with two graphs or two equations that are not
lines.
The blue one is going to end up being a parabola
and the green one I know that
is a line because it's of degree one there's no exponent.
Okay, so we should have...well,we don't know how many points of intersection
we'll have. We could have more than one though is the point.
So let's begin
by solving one of the two equations, either one,
for one of its variables, it doesn't really matter which,
the green equation looks easier
and probably I'd solve for x. But I wanna show you that there's no
really wrong answer here. I mean as far as which to solve for. So this time
I'm gonna solve for y, just from habit. I'm used to solving for Y.
So let's see; I'll subtract x from both sides
and then
that's a negative y. I don't want negative y; I want y so I'll divide everybody by -1.
And will equal x minus one; that's the first step to solve for one of the
variables.
Step b
is to substitute the expression equivalent to that variable
which in this case is y, the expression
equivalent to that is going to be X -1 and I'm gonna substitute it into the blue
equation
instead of y
I'm gonna put what y is equal to,
X-1. Now that creates an equation
that I should be able to solve.
Well, this is a quadratic equation
because it has an x squared but we know how to solve a quadratic
equation.
But we also know that we're probably gonna get 2 answers.
Let's do it. I distribute that -3; I get x squared -3x plus three.
Recall a negative times a negative equals a positive. Now, let's
begin
by getting a 0 on one side. I'll subtract one from both sides
and I'll get X squared minus 3x plus 2 equals 0.
Now remember one way to solve a quadratic
if you can do it, is to use
Okay, so let's factor and I see that there are two answers here
using the zero property
What do we make X equal to make this factor 0?
That's going to be a root.
One of the answers is that x equals one.
Now, in this one, the other answer will be what makes that equal to 0.
X equals two. Now these are both x's
so it looks like I'm gonna have 2 points of intersection.
One of the x's will be one and the other x will be two.
Now I need to substitute; I'll do it separately. Each of those X's
into the other equation and find y. I don't understand.
Well, since I have two points of intersection I have to do two
separate substitutions. So I'll put one in instead x
and solve for y
subtracting one
I get negative y equal 0. If I divide by -1, I get
y is zero. So in that case
I end up with the point one for x
and 0 for y that's one point of intersection. I put a two in for x
and solve for y.
Let's say subtract two from both sides
and then divide by -1.
I'll get y equals 1
so the other point will be X is one and
I'm sorry x is two and y is one 0.
There are actually two points of intersection between this parabola
and this line.
Ooop! Almost forgot the last step!
And, and since I've done twice the work I'm gonna have to check
2 answers into the other equations. Let's put (1,0) into this.
A one in for x and 0 in for y and I get,
son-of-a-gun, one does equal one so that one checks.
In the other point, I put (2.1). 2 in for x and one in for y.
Let's see, two squared minus 3
one does equal one; that one also checks.
So I'm in good shape. They checked. So these are the two points of intersection
Let's check in the original one also just to
make sure.
(1,0)
1-0 does equal one.
We're looking good and (2,1).
Well ,two -1, if we put that in, does equal one. So guess what
We did find the 2 points of intersection.
Well, we have to look at this graph just to see.
If we graphed this, course we'd have to solve for y first,
this parabola ,and that's what it would be if you
solved it for y.
and if we graphed it
it would look like this. Now the green line
course we'd have to solve that for y also it
would equal X-1 and if we graph that
going very I'm beyond that's where that would be
and, guess what, one point of intersection is the point (1,0)
and the other point of intersection
is in fact the point (2,1) so it does work. We didn't need to graph it
but it's nice to see what's going on isn't it?
So here's what we did in summary. We looked at the two equations.
We solved one of them for one of its variables,
doesn't matter, whichever is easiest and we substituted the expression that that
variable was equal to
or equivalent to into the
other equation and that created a now single variable equation
that we solved. When we solve it, we actually get a value for that variable
and then we substitute that value
into either of the equations and solve for the
other variable and then finally, very important after all that work
check your proposed
point. Make sure it works in both original equations.
Don't forget there's another way to check your
points of intersection that you found.
Your calculator!If you turn this baby on
and graphed or inserted the
functions that you put in. First we'll put y equals...
Get rid of something now that you used before
and put in that quadratic and (don't hit
graph yet) go down because remember you can graph up to seven of these things.
Hit the down cursor; clear out anything that may be in there
and then put in the other, the linear equation x-1.
Now you can graph and see the picture
but then you can also hit the trace button
and move left or right and move on to the points of intersection
and see in fact at least approximately that they are the points that you think
they are.
Okay? And hit clear to get out.
If you want to get a more accurate
answer for this point of intersection
you may wanna zoom in and and then trace along it
to see what the points of intersection are
or...that gives me an idea
You can let the calculator assist you in being even more exact and that is using
the CALC key.
Remember that second trace,
second trace yields CALC.
So if you have these functions already inserted into your calculator and you
hit second trace
choose number five, intersection,
and then it'll ask you what's the first curve
and you move on to that first curve but it'll probably jump onto one and hit
enter
and then move on to the other curve for the other
equation and hit enter and then
move near one of the points of intersection. That's why you need to make
a guess. It wants to know which point of intersection you want. So cursor over
very close to the point of intersection and hit enter and, son of a gun,
the calculator will tell you exactly where that point is
or at least out to as many decimal places as you have it set.
Okay, well, it's time to go do your homework so
get to it.