WEBVTT
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Hello, and welcome back
to a step in practice.
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We are dealing
today with equations
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that involve absolute values.
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In exercise 1, the
equation is absolute value
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of 3x minus 7 equals to minus 8.
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Well, it is clear
at the first glance
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that this equation
has no solutions.
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Why?
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Because the absolute
value of something
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is greater or equal
than 0, whereas minus 8
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is strictly negative.
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And so the set of
solutions is the empty set.
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Alternatively, one could have
used the method that Francis
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showed us in the last step.
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More precisely, the
method is that to consider
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two different cases, depending
on the sign of 3x minus 7.
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So alternatively,
if 3x minus 7 is
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greater or equal than 0, that
is x greater or equal than 7/3,
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the equation is equivalent to
3x minus 7 equal to minus 8,
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that is x equal to minus 1/3.
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However, minus 1/3-- well,
does not belong to the interval
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7/3 plus infinity.
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So the first set of
solutions is the empty set.
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The second case is 3x minus
7, strictly less than 0,
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that is, x strictly
less than 7/3.
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Well in that case the
equation is equivalent
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to minus 3x minus 7 equals
2 minus 8, which is minus 3x
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equals to minus 15,
which gives x equal 5.
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However, 5-- well,
it's bigger than 7/3,
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and so the second set of
solutions is the empty set.
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And the conclusion, again is
that the set of solutions which
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is the union of the
two sets of solutions,
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considering the two cases,
is again the empty set.
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So two different methods to
find the same conclusion.
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In exercise 2, we are dealing
with two absolute values.
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Well, the equation is
absolute value of 2x minus 4
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minus absolute value
of x minus 1 equal 3.
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We suggest to consider
different cases,
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depending on the sign of
2x minus 4 and x minus 1.
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Now, x minus 1 vanishes when
x equals 1, whereas 2x minus 4
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vanishes when x is equal to 2.
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So let us consider a table
with the values 1 and 2.
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And here, let us put the sign
of x minus 1 and the sign of 2x
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minus 4.
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Now, 2x minus 4 vanishes
in 2, whereas x minus 1
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vanishes in 1.
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Here we have the positive
sign, negative sign,
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here we have a negative sign,
and then a positive sign,
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Now we consider these
three different cases.
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If x is less than 1, then
both terms are negative.
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So the equation is equivalent
to minus 2x minus 4,
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minus and then again minus
x minus 1 equal to 3.
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And this is minus 2x plus 4
plus x minus 1 equal to 3,
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and this is again is
equivalent to minus x plus 3
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equal to 3, which
gives x equal to 0.
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0 less than 1.
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So we get a set of solution 0.
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Now, the second case is x
greater than 1 and less than 2.
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Well, in that case, x minus
1 is positive, 2x minus 4
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is negative, so the equation is
equivalent to minus 2x minus 4,
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which is the absolute value
of 2x minus 4, and x minus 1
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is positive.
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So minus x minus 1
equal to 3, and this
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is equivalent to
minus 3x plus 4 plus 1
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to this plus 5 equal to 3,
which gives 3x equals to 2,
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and then x equal to 2/3.
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Now, 2/3 is less
than 1, and thus we
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get a second set of solutions
which is the empty set.
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Finally, the third case,
x greater or equal than 2.
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Well, every term is positive.
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So the equation is equivalent
to 2x minus 4 minus x minus 1
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equal to 3.
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But this is also x minus 4
plus 1, so minus 3 equal to 3,
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and this is equivalent
to x equal to 6.
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Well, let us look at
the compatibility--
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6 is greater than
2, and thus we get 6
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as a third set of solutions.
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Now, the set of
solutions to our equality
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is the union of the
three sets that we found
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in each of the different cases.
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And so we get 0 and 6, and this
is the solution of exercise 2.
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In exercise 3, we consider
the equation absolute value
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of x minus 1 equals
to minus 4 plus 2x.
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So we can split this
equality into two equations
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by following what
Francis showed us.
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So in case where x minus 1 is
greater or equal than 0, that
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is x greater or equal than 1, we
get x minus 1 equal to minus 4
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plus 2x, that is, x
equals minus 1 plus 4, 3.
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So 3 is greater or equal than 1.
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And so we have a first
solution, which is 3.
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Let us check--
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3 minus 1 is 2.
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The absolute value of 2 is
2, and minus 4 plus 6 is 2.
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So we get the equality.
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If x minus 1 is less than 0,
that is, x is less than 1,
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and we take care
of this condition,
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the equality is equivalent
to minus x minus 1
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equals to minus 4 plus
2x, which is 3x equal to 1
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plus 4, 5, that
is x equal to 5/3.
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However, 5/3 is
greater than 1, so we
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have to exclude the solution.
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And the second set of
solutions is the empty set.
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So finally, the set
of solutions is just
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S1 that is the singleton 3.
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And this ends this
step in practice.
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And in the next
step in practice,
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you meet again Alberto.
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Be careful.
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He is a little bit
too a fan of algebra.
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See you next week.
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Bye.
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