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Solve the system of linear
equations by graphing, and they give us
two equations here. 5x plus 3y is equal to 7, and
3x minus 2y is equal to 8. When they say, "Solve the
system of linear equations," they're really just
saying find an x and a y that satisfies
both of these equations. And when they say to
do it by graphing, we're essentially going to
graph this first equation. Remember, the graph is
really just depicting all of the x's and y's that
satisfy this first equation, and then we graph
the second equation that's depicting all of the x's
and y's that satisfy that one. So if we were looking for an
x and a y that satisfies both, that point needs to
be on both equations or it has to be on both graphs. So it'll be the intersection
of the two graphs. So let's try to see
if we can do that. So let's focus on
this first equation, and I want to graph it. So I have 5x plus
3y is equal to 7. There's a couple of ways
we could graph this. We could put this in
slope-intercept form, or we could just pick
some points here. You just really need two
points to graph a line. So let me just set
some points over here. Let's say x and y. When x is equal to 0, what
does y need to be equal to? So when x is equal
to 0, we have-- let me do it over here-- we
have 5 times 0, plus 3 times y, is equal to 7. That's just 0 over there. So you have 3y is equal to 7. Divide both sides
by 3, you get y is equal to 7/3, which is
the same thing as 2 and 1/3 if we want to write
it as a mixed number. Now let's set y equal to 0. So if we set y as equal to
0, we get 5x, plus 3 times 0, is equal to 7. Or in this part right over
here, it just becomes 0. So we have 5x is equal to 7. Divide both sides
by 5, and we get x equals 7/5, which is the
same thing as 1 and 2/5. So let's graph both
of these points, and then we should be able to
graph this line, or at least a pretty good
approximation of that line. So we have the
point, 0, 2 and 1/3. So that's that point
right over there. So I'll call it 0,
7/3 right over there, and then we have the point,
7/5, 0, or 1 and 2/5, 0. So 1 and 2/5. 2/5 is a little
less than a half. So 1 and 2/5, 0. So our line is going to
look something like this. I just have to connect the dots. It's always hard to
draw the straight line. I'll draw it as a dotted line. So it would look
something like this. Normally, when you have to
solve a system of equations by graphing, they normally
give you a little bit cleaner numbers, but we'll
try our best and see if we can see where these
two lines intersect. So now, let's worry about the
second one right over there. So we have 3x minus
2y is equal to 8. So I'll do the same thing. So 3x minus 2y is equal to 8. We'll just look at the
x- and the y-intercepts. So first, the y-intercept. When x is equal to 0, this whole
thing boils down to 3 times 0 minus 2y is equal to 8. That's just 0. So you have negative
2y is equal to 8. Divide both sides
by negative 2, we get y is equal to negative 4. So the y-intercept
is 0, negative 4. Right over here, and we
mark it 0, negative 4. And then, let's
set y equal to 0. So when y is equal to 0,
this term right over here just becomes 0. So we get 3x minus 2
times 0, so that's just 0. So 3x is equal to 8. Divide both sides by 3,
you get x is equal to 8/3. And 8/3 is the same
thing as 2 and 2/3. So it puts it right about there. That's the point, 0, 8/3. Now let me try my
best to graph it. Connect these two dots. So let me do my very best. I drew a dotted line there. It goes something like that. And just eyeballing it, it looks
like these two lines intersect right over there. I'm hoping that this will
give us a clean answer. And this is the
point 2, negative 1. So that is the
point, 2, negative 1. Right? The x value here is 2.
y value is negative 1. Now, that's what we got
just by eyeballing it, and clearly these are
hand-drawn graphs. Not very precise. Let's verify or let's
see if 2, negative 1 does satisfy both
of these equations, if it's an x and y value
that satisfies both and lies on both the graphs. So if you put 2, negative
1 in this first equation, you get 5 times 2, plus
3 times negative 1, and we're going to see
if that is equal to 7. So this is 10 plus negative 3. So that's the same
thing as 10 minus 3. Does that equal 7? And yeah, it does. 10 minus 3 is equal to 7. So 2, negative 1 is
definitely on that graph or definitely satisfies
that equation. And then, let's do it
with the other one. So if we do 2, negative
1, you have 3 times 2, minus 2 times negative 1. And we're testing to see
if that is equal to 8. So 3 times 2. 3 times 2 is 6. And then 2 times
negative 1 is negative 2. But then, we're subtracting
that, so it's 6 plus 2. 6 plus 2 is equal to 8. And it definitely does equal 8. So we have the coordinate
or we have the point, 2, negative 1 satisfying
both equations. So we've solved the system of
linear equations by graphing.