Absolute Value Equations and Inequalities
Lesson 11
Lesson 11 is absolute value equations and inequalities, our warm up will begin with a
review of two step equations and solving and graphing a two step inequality. We will
being lesson 11 with our SOLVE problem; All wrestlers in the light weight division
must weigh within in 7 pounds of 140. The equation the absolute value of X minus
140 equals 7, represents the given situation. What are the minimum and maximum
weights? We will begin by our underlining the question, for “S” which is studying the
problem, our question which is what are the minimum and maximum weights? The
second step in “S” is to answer the question what is this problem asking me to find.
This problem is asking me to find the minimum and maximum weights.
We will being the lesson by discussing absolute value, we know that a distance
always has to be positive, for example no one could ever walk a negative 200 feet or
drive a negative 25 miles, you would just be driving in an opposite direction. Our
definition for the absolute value is, the distance from zero. Absolute value is the
distance from zero. Therefore the absolute value of any number must be positive
because the distance is always positive.
If we are going to find the absolute value of 5, we will go to the number 5 on the
number line and count how many units it is to zero, 1, 2, 3, 4, 5, so the absolute
value of 5 is 5 because the distance is 5, 1,2, 3, 4, 5 units. If we want to find the
absolute value of negative 5, we will find negative 5 on the number line and count to
see how many units negative 5 is from zero. Remember that the absolute value is
the distance from zero, 1, 2, 3, 4, 5; so the absolute value of negative 5 is also equal
to 5. 5 is a distance of 5 units from zero, and negative 5 is also a distance of 5 from
zero. So the absolute value of 5 is 5 and the absolute value of negative 5 is also 5.
When solving absolute value equations, we will actually create two separate
equations. Our first equations will be X equals 5. It is the positive value of the
expression inside the absolute value our second equations is the opposite of X
equals 5. Our second equations is the opposite of the expression inside the absolute
value equations. Our first equations in this case is already solved because X is
isolated, so our first value for X is 5. In our second equation we have a negative X
equals 5; in order to isolate X we must divide each side by negative 1. Leaving us
with the answer X equals negative 5. We have two answers, X equals 5 and X equals
negative 5. We can check our answers by plugging them back into our original
equation. Our original equation is the absolute value of X equals 5. If we plug our
first answer of 5 in, the absolute value of 5 equals 5; and this is a true statement
because the absolute value of 5 is 5, it is 5 units from zero. Where the distance from
zero is 5; so 5 equals 5. We must check our second answer as well, we will plug in
negative 5 for X, Is the absolute value of negative 5, 5? Yes, it is, because the
distance from zero is 5. 5 equals 5 so both of our answers check.
When you have an equation where the absolute value expression is equal to a
negative, your answer is automatically no solution. This is telling us that the absolute
value is equal to negative 5, the absolute value of any number or expression can
never be equal to a negative because that would be that a distance was negative
and we have already said that the absolute value of anything has to be a positive
number because the distance can never be negative. You will never walk a negative
50 feet. So anytime an absolute value expression is isolated and equal to a negative
your answer is no solution. To solve problem one we will create our two equations.
Our first equation will be the positive X plus 3 equal to 5 and our second equation will
be the opposite of what’s in side the absolute value bars, so we must us parenthesis,
to make sure that we distribute the negative to both terms. If we solve our first
equation we want to isolate X. we have a plus 3 on the same side as X, so the
opposite of plus 3 is subtract 3 or minus 3, keeping our equation balanced we will
subtract 3 from both sides, and X will equal 2. Our second equation, our first step is
to distribute the negative, we must distribute the negative to both terms, leaving us
with negative X minus 3 equals 5. From here we will need to add 3 to both sides in
order to isolate the negative X, the opposite of subtract 3 is to add 3, this will leave
us with negative X equal to 5 plus 3 which is 8. Our X is still not isolated so we must
divide both sides by negative 1, a negative divided by a negative is a positive X equal
to negative 8. Our two answers are X equals 2 and X equals negative 8. From here
we will plug both answer back into the original equation to check our answers. Write
the original equation, the absolute value of X plus 3 equals 5; if we plug in our first
answer the absolute value of 2 plus 3 equals 5; 2 plus 3 is 5 and the absolute value
of 5 is 5. so our answer checks. If we want to check our second equation, we write
our original equation and plug in our answer of negative 8 for X, negative 8 plus 3, is
the absolute value of negative 5, the absolute value of negative 5 is 5, because there
is a distance of 5 from zero, so our second answer checks.
In problem one our first step is to isolate the absolute value expression, we cannot
create our two equations until the absolute value expression is isolated on one side.
So in order to do this we have the absolute value of X plus 5, we want to isolate the
absolute value of X, so the opposite of plus 5 is subtract 5 or minus 5, we must
subtract 5 from both sides in order to isolate the absolute value of X, and 9 minus 5
is 4. we now have the absolute value of X isolated, so now we can create our two
equations. Our first equation is a positive of what’s inside the absolute value
expression equal to 4, and our second equations is the opposite, of what’s inside the
absolute value expression, equal to 4. We don’t have to solve anything on our first
equation because X is already isolated. On our second equation we have negative X
or negative 1 times X, which gives us negative X equal to 4, and in order to isolate the
X we are going to divide both sides by negative 1, X equals negative 4. so our 2
answers are X equals 4 and X equals negative 4. Now we are going to check our
answers. We will first write our original equation, the absolute value of X plus 5
equals 9. If we plug 4 in for X, we have the absolute value of 4 plus 5 equals 9, the
absolute value of 4 is 4, so 4 plus 5 equals 9 and 4 plus 5 is 9 so our answer checks.
We will check our second answer by plugging into the original equation, the absolute
value of negative 4 plus 5 equals 9, the absolute value of negative 4 is 4, and 4 plus
5 is equal to 9, so our second answer checks. We have two answers, X equals 4 and
X equals negative 4. In problem two our absolute value expression is not isolated,
so we must first isolate our absolute value expression, in order to do this we have a
multiply by 5 next to the absolute value expression, so we must divide each side by 5,
5 divided by 5 is 1, they cancel which leaves us with the absolute value of X; and 40
divided by 5 is 8. We will create our two equations, where X equals 8 or the positive
of what’s inside the absolute value expression, and our second is the opposite of X
equals 8. If X equals 8 the X is isolated so we do not need to do anything further. We
must multiply the X by negative 1 leaving us with negative X equals 8. In order for us
to isolate the 8 we must divide both sides by negative 1, leaving us with X equals
negative 8. We now have two answers that we need to check, to check we will begin
by writing down the original equation, we have 5 times the absolute value of 8 equals
40, 5 times, and the absolute value of 8 is 8, and 5 times 8 is 40, so our first answer
checks. To check our second answer we will write down our original equation, and
plug in negative 8 for X it is 5 times the absolute value of negative 8, and the
absolute value of negative 8 is 8, and 5 times 8 is 40 and 40 equals 40, our second
answer checks. So we have one answer of X equals 8 and another answer of X
equals negative 8.
We will solve our absolute value inequality using the same method we used with
equations. The only different is instead of having an equal to sign we now have a
less than or equal to sign. This will give us an inequality for, two inequalities as our
answer. And we will be able to graph our solution on the number line. The first
problem the absolute value of X plus 4 is less than or equal to 7, is a less than
problem. If the problem is less than, you hear the word and at the end, so we know
our two equations are X plus 4 is less than or equals to 7 and the opposite of X plus
4 is less than or equal to 7. To solve our first inequality we must isolate the X, we
have a plus 4 and the opposite of plus 4 is subtract 4, we see that our answer will be
X is less than or equal to 3. In our second equations we must first distribute the
negative to both terms inside the expression, we will have a negative X minus 4 is
less than or equal to 7, from here we want to isolate the X so we must add 4 to both
sides, the opposite of add 4 is subtract 4, leaving us with negative X less than or
equal to 11, and we still have a negative X so we will divide both sides by negative 1.
When you divide by a negative in inequalities you must flip your inequality sign. So
now we have X is greater than or equal negative 11, and X is less than or equal to 3.
When we graph our solution we have less than or equal to so we will have a solid dot
on 3 and we have greater than or equal to negative 11 so we will also have a solid
dot on negative 11. Since there is an and it is greater than negative 11 and less
than 3, an and will have a bar bell looking graph. We can choose any number in our
solution to check and make sure we have the correct solution. Suppose we want to
try zero, zero is in the solution. If we want to check we rewrite the original inequality,
which is the absolute value of X plus 4 is less than or equal to 7, if we plug zero in we
have zero plus 4 is less than or equal to 7, zero plus 4 is 4, and the absolute value of
4 is 4. So we ask ourselves, is 4 less than 7? Yes, it is so zero is a correct solution
and this is our correct graph. We know that X less than or equal to 3 AND X is greater
than or equal to negative 11.
In problem 2, we have the inequality the absolute value of 2X is greater than 10,
because it is greater we see that we will have two equations and our answer will be
connected with an OR. Because greater ends in the word OR. Our first inequality will
be 2X is greater than 10. and our second inequality will be negative 2X, or the
opposite of 2X, is greater than 10. In order to isolate the X in the first inequality we
will divide both sides by 2, leaving us with X is greater than 5. In the second
inequality we will multiply or find the opposite of 2X, so it’s negative 2X is greater
than 10, in order to isolate X we will divide both sides by negative 2, and when you
divide by a negative in inequalities you must flip your inequality sign, from greater
than to less than, and 10 divided by negative 2 is negative 5. When we graph our
solution we must use open circles on 5 and negative 5 because we have greater than
and less than. To graph the first inequality of X is greater than 5, we will draw an
arrow to the right because all of the numbers to the right are greater than 5. To
graph the solution X is less than negative 5, we will draw an arrow to the left,
because all of the numbers to the left are less than negative 5. If we want to check
our answer we write the original inequality, the absolute value of 2X is greater than
10. We can choose any number in our solution which is less than negative 5 or
greater than 5. After negative 5 the next integer is negative 6, let’s try negative 6 to
make sure it is a solution to our inequality. If I plug in negative 6 for X we see that 2
times negative 6 is negative 12, so we have the absolute value of negative 12 is
greater than 10. The absolute value of negative 12 is 12 and 12 is greater than 10.
So we have a correct solution. Our final answer is X is greater than 5 or X is less than
negative 5.
We will now go back to our SOLVE question from the beginning of the lesson, we
already have studied the problem and we know that this problem is asking us to find
the minimum and maximum weights. In “O” we are going to organize the facts, first
we have to identify the facts. All wrestlers in the light weight division must weigh
within 7 ponds of 140, first fact; the equation of the absolute value of X minus 140
equals 7 represents the given situation is our second fact, our first fact we will record
it in “O”. Our second fact is also important so we will write the equation. In “L” we
will line up a plan, we will choose our operation or operations and line up our plan by
writing what our plan of action will be. We will use addition on one equation and
multiplication and subtraction on one equation. Our plan will be to create 2
equations and solve them using opposite operations. In “V” we will verify our plan
with action, our first step in “V” is to estimate. We know that one of our weights will
be greater than 140 and one will be less than 140, so this will be our estimate. To
carry out our plan we will create our two equations from the equation that was given
in the problem. The absolute value of X minus 140 equals 7. Our first equation will
be the positive X minus140 equals 7 and our second equation will be the opposite of
X minus 140 equals 7. We will solve our first equation using addition we will add 140
to both sides, X will equal 147. In our second equation we must first distribute the
negative to both terms, so we have negative X plus 140 equals 7; we have a plus
140 so we will subtract 140 from both sides, leaving us with negative X equals
negative 133. We will divide both sides by negative 1 to isolate the X; and X equals
133. In “E” we will examine our results. The first questions we ask is does our
answer make sense, if we go back to our question what are the minimum and
maximum weights, 147 and 133 could be weights. The second to ask is, is your
answer reasonable, we go to our estimate and our estimate says, one will be greater
than 140, 147, and one will be less than 140, 133 is less than 140. We also have to
ask ourselves, is our answer accurate? We can check our answer by plugging into
the original absolute value equation. Our absolute value of X minus 140 equals 7.
We write the original equation and we plug in 147. The absolute value of 147-140 is
equal to the absolute value 7 equals 7; and the absolute value of 7 is 7. so our first
answer checks We will write our original equation to check our second answer we
have 133 minus 140 equals 7, the absolute value of 133 minus 140 is the absolute
value of negative 7 equals 7; and the absolute value of negative 7 is 7. Both of our
answers check so now we will write our final answer as a complete sentence. The
minimum weight is 133 and the maximum weight is 147.
We will close our lesson by reviewing the essential questions. Our first questions is,
does every number have an absolute value? Yes! The absolute value of all positive
numbers is itself. The absolute value of all negative numbers is its opposite, making
it a positive number, and the absolute value of zero is zero. A distance must always
be positive so the absolute value of any number is positive; except for zero because
the distance from zero is zero. Question 2, how many solutions do absolute value
equations have? Absolute value equations can have 1 solution, 2 solutions, or no
real solution. Question 3, how many solutions do absolute value inequalities have?
Absolute value inequalities have infinite many solutions there points or many
solutions that we could plug in that will make the inequality true.