SOLUTION: How do you solve and graph: /c-4/ > 1 The c-4 is in absolute value lines. Thank you!

Question 73747: How do you solve and graph:
/c-4/ > 1
The c-4 is in absolute value lines.
Thank you!
Found 2 solutions by jim_thompson5910, bucky:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Since the absolute value of a positive and a negative number is that same number but always positive, you can write an absolute value expression like as this and So this problem can be written as
Now we can solve for c

Which finally looks like

For the graph you would shade above y=1 and below (we let c=x). So shade in between the green and the red lines; however, don't shade in the middle triangle, since the is not greater than 1 in this region.
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Given:
.

.
solve and graph
.
One way to do absolute value problems is to solve them as two separate problems. For the
first problem, remove the absolute value signs and solve:
.
+(c-4)>1
.
You can treat this just as you would and equation ... except for one difference. That difference
is that if you have to multiply or divide both sides by a negative number, you must then
reverse the direction of the inequality sign. Don't forget that.
.
Now back to solving:
.
+(c-4)>1
.
The parentheses are preceded by a plus sign so you can just remove them and the inequality
becomes:
.
+c - 4 > 1
.
Just as you would in an equation, eliminate the -4 on the left side by adding 4 to both
sides to get:
.
+c > 5
.
That's the first limit on the value of c ... it must be greater than +5.
.
Now, on to the second equation. We do the same thing as before ... remove the absolute
value signs, except this time we put a minus sign in front of the quantity that was inside
the absolute value signs. When you do this you have:
.
-(c-4)>1
.
Because the quantity is preceded by a negative sign, when you remove the parentheses
you must change the sign of every term that was inside the parentheses. When you do
that the inequality becomes:
.
-c + 4 >1
.
We need to solve this for +c. Let's begin by eliminating the +4 on the left side by
subtracting 4 from both sides. This causes the inequality to become:
.
-c > -3
.
To change the equation so that the left side is +c, multiply both sides by negative 1.
And when you do, do NOT forget to reverse the direction of the inequality sign to get:
.
+c < +3
.
This tells us that c must be less than +3
.
Combined, the two restrictions are that c can be any number less than 3 and it can also
be any number greater than 5. But it cannot be either 3 or 5 or any number between those
two values.
.
Let's check that out with a trio of trials. Start with the problem:
.

.
Let's set c equal to zero. That's a number less than +3 so it should work. When you substitute
0 for c you get:
.
which simplifies to:
.
and since we get +4 > 1. That works!
.
Now let's set c equal to 4. That should NOT work. Substitute 4 for c and get:
.
this obviously becomes:
.
and the absolute value of 0 is 0. Obviously 0 > 1 is NOT true so numbers
between 3 and 4 probably all need to be excluded.
.
Finally try c = 6. That is greater than +5 and should work. Substitute +5 for c and get:
.
which becomes and simplifies to .
At this point it is obvious that 2 > 1 is true and that adds to the likelihood that numbers
greater than 5 will work.
.
We have a good solution!
.
To graph, just make a number line and put dots at +3 and +5. Exclude those dots and shade
the number line from just to the left of +3 all the way to the left and from just to
the right of +5 all the way to the right. c can take any value in the shaded region.
.
Hope this shows you a way to work absolute value inequalities. Remember the two different
equations to solve (one by using the absolute value quantity with a + sign and the other
by using that quantity with a minus sign preceding it), the need to reverse the direction of
the inequality sign if multiplying or dividing both sides by a negative number, and to
always solve for the positive value of the variable. The rest is just algebraic manipulation.
.
Hope this works for you.
.

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