All Precalculus Resources
Example Questions
Example Question #4 : Solve Absolute Value Inequalities
Solve for .
Our first step in solving this equation is to isolate the absolute value. We do this by dividing both sides by
.
We then set up our two equations:
and .
Subtracting 4 from both sides leaves us with
and .
Lastly, we multiply both sides by 2, leaving us with :
and .
Which can be rewritten as:
Example Question #3 : Solve Absolute Value Inequalities
Solve for
We first need to isolate the absolute value, which we can do in two steps:
1. Add 2 to both sides:
2. Divide both sides by 4:
Our next step is to set up our two equations:
and
We can now solve the equations for by subtracting both sides by 8:
and
and then dividing them by 5:
and
Which can be rewritten as:
Example Question #1 : Solve Absolute Value Inequalities
Solve the following absolute value inequality:
First we need to get the expression with the absolute value sign by itself on one side of the inequality. We can do this by subtracting seven from both sides.
Next we need to set up two inequalities since the absolute value sign will make both a negative value and a positive value positive.
From here, subtract thirteen from both sides and then divide everything by four.
Example Question #8 : Linear Inequalities
Solve the following absolute value inequality:
First we need to get the expression with the absolute value sign by itself on one side of the inequality. We can do this by dividing both sides by three.
We now have two equations:
and
So, our solution is
Example Question #1 : Solve Absolute Value Inequalities
Solve the following inequality:
First we need to get the expression with the absolute value sign by itself on one side of the inequality. We can do this by subtracting two from both sides then dividing everything by three.
Since absolute value signs make both negative and positive values positive we need to set up a double inequality.
Now to solve for subtract four from each side.
Example Question #7 : Linear Inequalities
Solve for :
If , then either or based on the meaning of the absolute value function. We have to solve for both cases.
a) subtract 5 from both sides
divide by -2, which will flip the direction of the inequality
Even if we didn't know the rule about flipping the inequality, this answer makes sense - for example, , and .
b) subtract 5 from both sides
divide by -2, once again flipping the direction of the inequality
Example Question #1 : Solve Absolute Value Inequalities
Solve the absolute value inequality.
First, simplify so that the absolute value function is by itself on one side of the inequality.
.
Note that the symbol flipps when you divide both sides by .
Next, the two inequalities that result after removing the absolute value symbols are
and .
When you simplify the two inequalities, you get
and .
Thus, the solution is
.
Example Question #841 : Pre Calculus
To solve absolute value inequalities, you have to write it two different ways. But first, divide out the 4 on both sides so that there is just the absolute value on the left side. Then, write it normally, as you see it: and then flip the side and make the right side negative: . Then, solve each one. Your answers are and .
Example Question #842 : Pre Calculus
Solve the following:
To solve absolute value inequalities create two functions.
Simply remember that when solving inequalities with absolute values, you keep one the same and then flip the sign and inequality on the other.
Thus,
Then, you must write it in interval notation.
Thus,
Example Question #1 : Evaluate Expressions That Include The Inverse Tangent, Cosecant, Secant, Or Cotangent Function
Approximate:
:
There is a restriction for the range of the inverse tangent function from .
The inverse tangent of a value asks for the angle where the coordinate lies on the unit circle under the condition that . For this to be valid on the unit circle, the must be very close to 1, with an value also very close to zero, but cannot equal to zero since would be undefined.
The point is located on the unit circle when , but is invalid due to the existent asymptote at this angle.
An example of a point very close to that will yield can be written as:
Therefore, the approximated rounded value of is .