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## Алгебра 1

### Course: Алгебра 1 > Unit 6

Lesson 1: Теңдеулер жүйесіне кіріспе- Теңдеулер жүйесі: трольдер, ақылы төлемдер (2-ден 1)
- Теңдеулер жүйесі: трольдер, ақылы төлемдер (2-ден 2)
- Теңдеулер жүйесінің шешімін тексеру
- Теңдеулер жүйесінің шешімдері
- Графигімен берілген теңдеулер жүйесі: y=7/5x-5 & y=3/5x-1
- Графигімен берілген теңдеулер жүйесі: нақты және жуықталған шешімдер
- Графигімен берілген теңдеулер жүйесі
- Берілген контекст бойынша жүйе құру.

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# Графигімен берілген теңдеулер жүйесі: нақты және жуықталған шешімдер

Біз екі сызықты теңдеуден тұратын теңдеулер жүйесін стандартты түрде шешіп, кейін жауабы анық көрінбеген теңдеулер жүйесінің шешімін

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## Видео транскрипті

- The following two equations
form a linear system. This is one equation; it has X and Y so it's gonna define a line. And then I have another
equation that involves X and Y, so it's gonna define another line. It says: "Graph the system of equations "and find its solution." So we're gonna try to find it visually. So let's graph this first one. To graph this line, I have
the little graphing tool here. Notice if I can figure out two points, I can move those points around and it's going to define our line for us. I'm gonna pick two X values and figure out the corresponding Y values
and then graph the line. So let's see how I could do this. So let's see; an easy one is what happens when X is equal to zero? Well if X is equal to zero,
everything I just shaded goes away and we're left
with -3y is equal to nine. So -3y equals nine. Y
would be negative three. So when X is equal to zero,
Y would be negative three. So let me graph that. When X is equal to zero, X is zero, Y is negative three. Now another easy point
actually instead of trying another X value, let's think
about when Y is equal to zero 'cause these equations
are in a standard form so it's easy to just test. Well what are the X and Y intercepts? So when Y is equal to zero, this term goes away,
and you have negative X is equal to nine, or X would
be equal to negative nine. So when Y is zero, X is negative nine. So when Y is zero, X is negative nine, or when X is negative nine, Y is zero. So I've just plotted this first equation. So now let's do the second
one. We'll do the same thing. What happens when X is equal to zero? When X is equal to zero, so this is going to be
our Y intercept now. When X is equal to zero, -6y
is equal to negative six. Well Y would have to be equal to one. So when X is zero, Y is equal to one. So when X is zero, Y is equal to one. Get one more point here. When Y is zero, when this term is zero, Y being zero would make
this entire term zero, then 6x is equal to negative six or X is equal to negative one. So when Y is zero, X is negative one or when X is negative one, Y is zero. When X is negative one, Y is zero. And so just like that,
I've plotted the two lines. And the solution to the
system are the X and Y values that satisfy both equations;
and if they satisfy both equations, that means
they sit on both lines. And so in order to be on both lines, they're going to be at
the point of intersection. And I see this point of
intersection right over here, it looks pretty clear that
this is the point X is equal to negative three and Y
is equal to negative two. So it's the point negative
three comma negative two. So let me write that down. Negative three comma negative two. And then I could check
my answer; got it right. Let's do another. Let's
do another one of these. Maybe of a different type. So over here it says: "A
system of two linear equations "is graphed below. "Approximate the solution of the system." Alright so here I just have
to just look at this carefully and think about where this point is. So let's think about first its X value. So its X value, it's about right there
in terms of its X value. It looks like, so this is negative one. This is negative two, so negative 1.5 is gonna be right over here. It's a little bit to the
left of negative 1.5, so it's even more negative,
I would say negative 1.6. And I'm approximating it, negative 1.6. Hopefully it has a little leeway in how it checks the answer. What about the Y value? So if I look at the Y value here, it looks like it's a little
less than one and a half. One and a half would be
halfway between one and two. It looks like it's a little
less than halfway between one and two, so I'd give it 1.4, positive 1.4. And let's check the answer,
see how we're doing. Yep, we got it right. Let's actually just do
one more for good measure. So this is another system. They've just written the equations in more of our slope intercept form. So let's see, Y is equal to
negative seven, X plus three. When X is equal to zero,
we have our Y intercept. Y is equal to three. So when X is equal to
zero, Y is equal to three. And then we see that our
slope is negative seven. When you increase X by one,
you decrease Y by seven. So when you increase X
by one, you decrease Y by one, two, three, four, five, six, and seven. When X goes from zero to one, Y went from three to negative four, it went down by seven,
so that's that first one. Now the second one: our Y intercept. When X is equal to zero,
Y is negative three, so let me graph that. When X is zero, Y is
equal to negative three. And then its slope is negative one. When X increases by
one, Y decreases by one. So the slope here is negative one. So when X increases by
one, Y decreases by one. And there you have it. You have
your point of intersection. You have the X-Y pair that
satisfies both equations. That is the point of intersection. It's gonna sit on both lines which is why it's the
point of intersection. And that's the point X equals one, Y is equal to negative four. So you have X equals one and Y is equal to negative four. And I can check my answer
and we got it right.