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Μάθημα 2: Solving systems with elimination- Systems of equations with elimination: 3t+4g=6 & -6t+g=6
- Συστήματα εξισώσεων με αντίθετους συντελεστές
- Systems of equations with elimination: x+2y=6 & 4x-2y=14
- Systems of equations with elimination: -3y+4x=11 & y+2x=13
- Systems of equations with elimination: 2x-y=14 & -6x+3y=-42
- Systems of equations with elimination: 4x-2y=5 & 2x-y=2.5
- Systems of equations with elimination: 6x-6y=-24 & -5x-5y=-60
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Systems of equations with elimination: 6x-6y=-24 & -5x-5y=-60
Sal solves the system of equations 6x - 6y = -24 and -5x - 5y = -60 using elimination. Δημιουργήθηκε από τον Σαλ Καν.
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We never know when we might have
to do a little bit more party planning. So it doesn't hurt to
have some practice. And that's what this
exercise is doing for us, is generating
problems so that we can try solving systems of
equations with elimination. And so in this first
problem, it says solve for x and y
using elimination. And then this is
what they have-- 6x minus 6y is equal
to negative 24. Negative 5x minus 5y is
equal to negative 60. So let me get my scratch
pad out to solve this. Let me rewrite it. So they gave us 6x minus
6y is equal to negative 24. And negative 5x minus 5y
is equal to negative 60. So what we have to
think about, and we saw this in several
of the other videos, is when we want to
eliminate a variable, we want to manipulate
these two equations. And if we were to add
the corresponding sides, that variable might disappear. So if we just added a
6x to a negative 5x, that's not going
to cancel it out. If this was a negative
6x, that would work out. Or if this was a positive
5x, that would work out. But this isn't exactly right. So if I want to
eliminate the x, I have to manipulate these
equations so that these two characters might cancel out. And one thing that
pops into my brain is it looks like
all of this stuff up here is divisible by
6, and all of this stuff down here is divisible by 5. And if we were to divide
all this stuff up here by 6, we'd be left with
an x over here. And if we were to divide
all this bottom stuff by 5, we'd be left with a
negative x right over here. And then they just
might cancel out. So let's try that out. Let's take this first equation. And we're going to
multiply both sides by 1/6. Or another way you
could think about it is we're dividing both sides by 6. And as long as we do the
same thing to both sides, the equation holds. The equality holds. So if you multiply everything
by 1/6, 6x times 1/6 is just going to be x. 6y times 1/6 is just y. So it's negative y. Negative 24 times
1/6 is negative 4. Or you could just view it
as negative 24 divided by 6 is negative 4. So this equation, the blue one,
we've simplified as x minus y is equal to negative 4. Let's do something similar
with the second one. Here we could say we're going to
multiply everything times 1/5. Or you could say that we're
dividing everything by 5. If we do that, negative
5x divided by 5 is just negative x. Negative 5y divided
by 5 is negative y. And then negative 60
divided by 5 is negative 12. And now, this looks
pretty interesting. If we add the two left-hand
sides-- and remember, we can keep the equality,
because we're essentially adding the same
thing to both sides. You can imagine we're starting
with the blue equation. And on the left-hand side,
we're adding negative x minus y. And on the right-hand side,
we're adding negative 12. But the second equation tells
us that those two things are equal. So we're doing
the same principle that we saw when we first
started looking to algebra, that you can maintain your
equality as long as you add the same thing. On the left-hand side,
we're going to add this. And on the right-hand side,
we're going to add this. But this second
equation tells us that those two things are equal. So we can maintain our equality. So let's do that. What do we get on
the left-hand side? Well, you have a positive
x and a negative x. They cancel out. That was the whole point behind
manipulating them in this way. And then you have negative y
minus y, which is negative 2y. And then on the right-hand
side, you have negative 4 minus 12, which is negative 16. And these are going to
be equal to each other. Once again, we're adding the
same thing to both sides. To solve for y, we can divide
both sides by negative 2. And we are left with y
is equal to positive 8. But we are not done yet. We want to go and substitute
back into one of the equations. And we can substitute
back into this one and to this one, or
this one and this one. The solutions need to satisfy
all of these essentially. This blue one is another way of
expressing this blue equation. This green equation
is another way of expressing this
green equation. So I'll go for whichever one
seems to be the simplest. And this one seems to be
pretty simple right over here. So let's take x
minus y-- we just solved that y would be positive
8-- is equal to negative 4. And now to solve for x, we just
have to add 8 to both sides. And we are left with,
on the left-hand side, negative 8 plus 8 cancels out. You're just left with an x. And negative 4 plus 8
is equal to positive 4. So you get x is equal
to 4, y is equal to 8. And you can verify
that it would work with either one of
these equations. 6 times 4 is 24 minus 6 times
8-- so it's 24 minus 48-- is, indeed, negative 24. Negative 5 times 4 is negative
20, minus negative 40, if y is equal to 8, does,
indeed, get you negative 60. So it works out
for both of these. And we can try it out by
inputting our answers. So x is 4, y is 8. So let's do that. So let me type this in. x is going to be equal to 4. y is going to be equal to 8. And let's check our answer. It is correct. Very good.