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Current time:0:00Total duration:8:30

Let's say I have the equation
y is equal to x plus 3. And I want to graph all of the
sets, all of the coordinates x comma y that satisfy this
equation right there. And we've done this
many times before. So we draw our axis, our axes. That's my y-axis. This is my x-axis. And this is already in
mx plus b form, or slope-intercept form. The y-intercept here is y
is equal to 3, and the slope here is 1. So this line is going
to look like this. We intersect at 0 comma
3-- 1, 2, 3. At 0 comma 3. And we have a slope of 1,
so every 1 we go to the right, we go up 1. So the line will look
something like that. It's a good enough
approximation. So the line will
look like this. And remember, when I'm drawing
a line, every point on this line is a solution
to this equation. Or it represents a pair
of x and y that satisfy this equation. So maybe when you take x is
equal to 5, you go to the line, and you're going to see,
gee, when x is equal to 5 on that line, y is equal
to 8 is a solution. And it's going to
sit on the line. So this represents the solution
set to this equation, all of the coordinates
that satisfy y is equal to x plus 3. Now let's say we have
another equation. Let's say we have an equation
y is equal to negative x plus 3. And we want to graph all of the
x and y pairs that satisfy this equation. Well, we can do the
same thing. This has a y-intercept also
at 3, right there. But its slope is negative 1. So it's going to look
something like this. Every time you move to the
right 1, you're going to move down 1. Or if you move to the right a
bunch, you're going to move down that same bunch. So that's what this equation
will look like. Every point on this line
represents a x and y pair that will satisfy this equation. Now, what if I were to ask you,
is there an x and y pair that satisfies both of
these equations? Is there a point or coordinate
that satisfies both equations? Well, think about it. Everything that satisfies this
first equation is on this green line right here, and
everything that satisfies this purple equation is on the
purple line right there. So what satisfies both? Well, if there's a point that's
on both lines, or essentially, a point of
intersection of the lines. So in this situation, this
point is on both lines. And that's actually
the y-intercept. So the point 0, 3 is on
both of these lines. So that coordinate pair, or that
x, y pair, must satisfy both equations. And you can try it out. When x is 0 here, 0 plus
3 is equal to 3. When x is 0 here, 0 plus
3 is equal to 3. It satisfies both of
these equations. So what we just did, in a
graphical way, is solve a system of equations. Let me write that down. And all that means is we
have several equations. Each of them constrain
our x's and y's. So in this case, the first one
is y is equal to x plus 3, and then the second one is y is
equal to negative x plus 3. This constrained it to a line
in the xy plane, this constrained our solution
set to another line in the xy plane. And if we want to know the x's
and y's that satisfy both of these, it's going to be the
intersection of those lines. So one way to solve these
systems of equations is to graph both lines, both
equations, and then look at their intersection. And that will be the solution
to both of these equations. In the next few videos, we're
going to see other ways to solve it, that are maybe more mathematical and less graphical. But I really want you to
understand the graphical nature of solving systems
of equations. Let's do another one. Let's say we have y is
equal to 3x minus 6. That's one of our equations. And let's say the other equation
is y is equal to negative x plus 6. And just like the last video,
let's graph both of these. I'll try to do it as
precisely as I can. There you go. Let me draw some. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. And then 1, 2, 3, 4,
5, 6, 7, 8, 9, 10. I should have just copied and
pasted some graph paper here, but I think this'll
do the job. So let's graph this purple
equation here. Y-intercept is negative 6, so
we have-- let me do another [? slash-- ?] 1, 2, 3, 4, 5, 6. So that's y is equal
to negative 6. And then the slope is 3. So every time you move
1, you go up 3. You moved to the right 1,
your run is 1, your rise is 1, 2, 3. That's 3, right? 1, 2, 3. So the equation, the line
will look like this. And it looks like I intersect
at the point 2 comma 0, which is right. 3 times 2 is 6, minus 6 is 0. So our line will look something like that right there. That's that line there. What about this line? Our y-intercept is plus 6. 1, 2, 3, 4, 5, 6. And our slope is negative 1. So every time we go 1 to the
right, we go down 1. And so this will intersect at--
well, when y is equal to 0, x is equal to 6. 1, 2, 3, 4, 5, 6. So right over there. So this line will
look like that. The graph, I want to get it
as exact as possible. And so we're going to ask
ourselves the same question. What is an x, y pair
that satisfies both of these equations? Well, you look at it here, it's
going to be this point. This point lies on both lines. And let's see if we can figure
out what that point is. Just eyeballing the graph here,
it looks like we're at 1, 2, 3 comma 1, 2, 3. It looks like this is the same
point right there, that this is the point 3 comma 3. I'm doing it just on inspecting
my hand-drawn graphs, so maybe it's
not the exact-- let's check this answer. Let's see if x is equal to
3, y equals 3 definitely satisfies both these
equations. So if we check it into the first
equation, you get 3 is equal to 3 times 3, minus 6. This is 9 minus 6, which
is indeed 3. So 3 comma 3 satisfies
the top equation. And let's see if it satisfies
the bottom equation. You get 3 is equal to negative
3 plus 6, and negative 3 plus 6 is indeed 3. So even with our hand-drawn
graph, we were able to inspect it and see that, yes, we were
able to come up with the point 3 comma 3, and that
does satisfy both of these equations. So we were able to solve this
system of equations. When we say system of equations,
we just mean many equations that have
many unknowns. They don't have to be, but they
tend to have more than one unknown. And you use each equation as a
constraint on your variables, and you try to find the
intersection of the equations to find a solution
to all of them. In the next few videos, we'll
see more algebraic ways of solving these than drawing their
two graphs and trying to find their intersection
points.