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## ತರಗತಿ 10 ಗಣಿತ (ಭಾರತ)

### Course: ತರಗತಿ 10 ಗಣಿತ (ಭಾರತ) > Unit 3

Lesson 3: Number of solutions algebraically# Forming systems of equations with different numbers of solutions

Sal write an equation that along with the equation 4x + 5y = 2 forms a system of equations with infinitely many solutions. ಸಾಲ್ ಖಾನ್ ರವರು ರಚಿಸಿದ್ದಾರೆ.

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## ವೀಡಿಯೊ ಪ್ರತಿಲಿಪಿ

Fill in the blanks to form
a system of linear equations in the variables x and y with
infinitely many solutions. So you're going to have
infinitely many solutions if essentially both
of these equations are describing the same line. If they're both essentially
the same equation, they are the same constraint. And we can graphically
imagine that. Let's say that that's our
y-axis, and this is our x-axis. And we could even try to
graph this right over here. If we were to put this
into slope-intercept form, you would have 5y is
equal to 2 minus 4x. And then if we divide both sides
by 5, you get y is equal to-- and I'll swap these--
negative 4/5 x plus 2/5. So this line up here is going
to look something like this. It's y-intercept is at 2/5. And it's going to have
a negative 4/5 slope. So it's going to look
something like this. So that's what the
line looks like. So we're going to get an
infinite number of solutions for the system. If the second line,
if when we graph it, is essentially the
exact same line, it overlaps at
every x and y that satisfy the first equation. Now, the easiest way to think
about what these blanks should be is, well, how do
I fill in the blanks here so it is really just a
direct algebraic manipulation of this first equation? And we have a clue here. On the right-hand side of the
first equation, we have a 2. On the right-hand side of the
second equation, we have a 4. So if we want a direct
algebraic manipulation, clearly on the right-hand side,
you have multiplied by 2. And if you want this to
be the same equation, you have to do
that to both sides. So let's multiply
both sides by 2. So 4x times 2 is 8x. 5y times 2 is 10y. So this equation
and this equation are the exact same constraint. They represent the
exact same line. They have an infinite
number of solutions. Let's do another example. Which of the following
choices of a will result in a system of linear
equations with no solutions? So you're going to have
no solutions as if you take the same
combination of x and y's, but you get to a
different number. Or another way of
thinking about it is if you plotted
this equation and it had the exact same slope, but
it had a different y-intercept. So let's think
about it both ways. So first, let's try to
algebraically manipulate the second equation so that
the left-hand side looks exactly the same as the
left-hand side up here. And then we just
have to make sure that the right-hand
side is different. And then you will
have no solutions. So let's think about
that a little bit. So it looks pretty close. If we multiply the left-hand
side by a negative 1, it's going to look just like
the left-hand side over here. So let's do that. Let's multiply the left-hand
and the right-hand side by negative 1. So it's essentially
the same equation. So negative 6x times
negative 1 is positive 6x. 7y times negative 1 is minus 7y. And a times negative
1 is negative a. So notice, now on the
left-hand side we have the same combinations of x's and y's. We have 6 x's minus 7 y's. If on the right-hand side these
two things were to be equal, then we'd have the
scenario that we just saw. We would have the
exact same line. If negative a was equal to 4,
we would have the same line. So let me write that down. If negative a is
equal to 4, then we are dealing with the same
line/equation/constraint. And like the
previous scenario, we would have an infinite
number of solutions. On the other hand, if
negative a is not equal to 4, then there's no way
that there's anything that will satisfy both of them. Here you're saying
that whatever your x is and whatever your y is, take
6 of the x's and subtract 7 of the y's, and you get 4. If here you take 6 of
the x's and subtract 7 of the y's and you get
a different number, then there's never
going to be an x and a y that satisfy
both of them. So if negative 8
is not equal to 4, then you're going to
have no solutions. Or another way of saying this is
if a does not equal negative 4, obviously, if a is
equal to negative 4, then negative a is
going to be equal to 4. So as long as a is not
equal to negative 4, you're going to
have no solutions. So a can be any number
except for negative 4. Now, another way
to think about this is to put them both in
slope-intercept form. And you'll see that they'll
have the same slope. And then you would
just want to have them have different
y-intercepts. And they're going to be parallel
lines that don't overlap. So this top one right over
here, if we subtract 6x from both sides,
you get negative 7y is equal to negative 6x plus 4. Divide both sides by negative 7. You get y is equal
to 6/7 x minus 4/7. So that's this first one
right up here written in slope-intercept form. Now, the second
one, I'll just start with exactly what they gave us. Let me add 6x to both sides. So this is the second one. So if I add 6x to both sides,
I get 7y is equal to 6x plus a. And then if I divide
both sides by 7, I get y is equal to
6/7 x plus a over 7. So notice in slope-intercept
form, our first equation looks like that. And in terms of a, our second
equation looks like that. So they definitely already
have the same slope. They have the exact same slope. If a is equal to
negative 4, then these are going to be the
exact same equations. They're going to have the
exact same y-intercept. You're going to have an
infinite number of solutions. On the other hand, if a is
anything other than negative 4, you're going to have a
different y-intercept. And these two things are
just going to be parallel.