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## Two-variable linear equations intro

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# Solutions to 2-variable equations: graphical (old)

## Video transcript

Is 3 comma negative 4 a
solution to the equation 5x plus 2y is equal to 7? So there's two ways
to think about it. One, you could just
substitute this x and y value into this equation to see if
it satisfies-- and then we'll do that way first--
and the other way is if you had a graph
of this equation, you could see if this point sits
on that graph, which would also mean that it is a
solution to this equation. So let's do it the first way. So we have 5x plus 2y is equal
to 7, so let's substitute. Instead of x, let
us put in 3 for x. So 5 times 3 plus 2 times
y-- so y is negative 4-- plus 2 times negative 4
needs to be equal to 7. I'll put a question mark
here, because we're not sure yet if it does. So 5 times 3 is 15, and then 2
times negative 4 is negative 8. So the left hand side, it
simplifies to 15 minus 8, and this needs to be equal to 7. And of course, 15 minus 8 does
equal 7, so this all works out. This is a solution, so
we've answered the question. But I also want to show
you, this way we just did it by substitution. If we had the graph
of this equation, we could also do it graphically. So let's give ourselves
the graph of this equation, and I'll do that by
setting up a table. There's multiple
ways to graph this. You could put it in a
slope-intercept form and all of the rest, but I'll just set
up a table of x and y values. And I'll graph it, and
then given the graph, I want to see if this
actually sits on it. And obviously it will,
because we've already shown that this works. In fact, we could try
the point 3, negative 4, and that actually
is on the graph. We could do it on our table,
but I won't do that just yet. I'm just going to do this
to give ourselves a graph. So let's try it when
x is equal to 0. We have 5 times 0 plus
2 times y is equal to 7. So when x is equal
to 0, y is going to be-- so you're going to
have 0 plus 2y is equal to 7. y is going to be equal to 3.5. When x is equal to 1, you
have 5 plus 2y is equal to 7. If you subtract 5
from both sides, you get 2y is equal to 2. You get y is equal to 1. So when x is 1, y is 1,
and when x is-- well let's try-- well that's actually
enough for us to graph. We could keep doing more points. We could even put the
point 3, negative 4 there, but let's just try to graph
it in this very rough sense right here. So let me draw my x-axis,
and then this right over here is my y-axis. And let me draw
some points here. So let's say that this
is y is 1, 2, 3, 4. This is negative 1, negative
2, negative 3, negative 4. I could keep going
down in that direction. This is 1. Let me do it a little
bit-- 1, 2, 3, 4, and I could just keep going on and
on in the positive x direction. So let's plot these points. I have 0, 3.5. When x is 0, y is 1, 2, 3.5. When x is 1, y is 1. And so if we were to draw
this line-- I'll do it as a dotted line, just
so that I can make sure I connect the dots. I can do a better
job than that though. So it will look
something like that. And so if someone
gave you this line, you'd say oh, well it's 3,
negative 4 on this line, and let's assume that
we drew it really nicely and this was all to scale. Let me try one
last attempt at it. So it's going to look
something like that. And If someone asked
is 3, negative 4 on it, you could visually do
it, but it's always hard when you actually
don't substitute it, because you don't know. Maybe you're a little bit off. But if you look at
it over here, you say when x is equal
to 3, what is y? Well, you go down
here, and it looks like y is equal to negative 4. So this is a point
3 comma negative 4. Obviously, in general,
you don't want to just rely off of
inspecting graphs. Maybe this was 3,
negative 3.9999, and you just couldn't
tell looking at the graph. That's why you always want
to just the substitute and make sure that
it really does equal, that this equality really
does hold true at that point, not just looking at the graph. But it's important to
realize that the graph really is another representation
to all of the solutions of this equation.