WEBVTT
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Hello.
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Let us consider
the first exercise
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of the "equations in several
variables in practice" step.
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We have to solve,
now, this system.
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You immediately realize that
this is not a linear system.
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Indeed, in the
second equation, it's
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not true that any segment is
a term of degree at most 1.
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For example, this
term has degree 3.
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When you have to solve
a system like this one,
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probably the more
reasonable method
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is the substitution method.
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Let us see how to
solve this system.
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You see, we have two equations.
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One is simpler than the other.
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The first one is
simpler, of course.
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And through the
first one, you easily
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can write y in function of x.
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Indeed, you have that y
is equal to 4 minus x.
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Then, now, we can substitute
in the second equation,
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in a more complicated one.
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We can substitute
to y 4 minus x.
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And you get x cubed plus
4 minus x cubed equal 28.
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And you see what happens, as
always with the substitution
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method, we get one equation
with only one variable.
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Let us try to find the
solutions of this equation.
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Good.
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Then we have x cubed plus.
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And we have to take the
cube of this binomial.
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And that is 4 to the cube.
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Then we have minus 3 times
the square of 4 times x,
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then we have plus
3 times 4 times
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the square of minus x, or
of x, which is the same.
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And finally, minus x cubed,
and this is to be equal to 28.
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Now, you immediately realize
that this term and this term
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can be erased, and we remain
with an equation of degree 2
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in x.
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And the equation
is the following.
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You have 12 times x
square, this term.
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Then we have this one.
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The square of 4 is 16 times
3, and you get minus 48.
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And then we have the constant
term, which is the cube of 4.
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Then you have to do 4 times
4, which is 16, times 4.
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16 times 4 is 64.
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And then you have this term,
which becomes minus 28,
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equals 0.
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64 minus 28 is equal to 36.
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You immediately realise that
all these coefficients are
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divisible by 12.
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Therefore, we can rewrite this
equation in a simpler way.
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Dividing by 12, you have x
square minus 4 times x plus 3
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equals 0.
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OK now we have an
equation of degree 2,
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and we compute its
discriminant, which
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is the square of minus 4 minus
4 times the leading coefficient,
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which is 1, times
the constant term.
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Of course, it is 16
minus 12, which is 4.
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The discriminant
is greater than 0.
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Therefore, this equation
will have two solutions.
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Which are the solutions?
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[They] are x equal
to minus minus 4,
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which is 4, plus or
minus the square root
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of the discriminant over 2
times the leading coefficient.
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Then we get 4 plus 2 divided
by 2, which is 3, or 4 minus 2
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divided by 2, which is 1.
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These are the two
possible solutions
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for our equation of
the second degree.
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Therefore, now, we can
obtain the values of y,
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remembering that y is
equal to 4 minus x.
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Then if x is equal to 3,
then y will be 4 minus 3,
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will be equal to 1.
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If x is equal to 1, then
y will be 4 minus 1, 3.
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Therefore, the
solutions of this system
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are the following pairs--
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the pair 3, 1,
and the pair 1, 3.
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Ciao a tutti.
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