# Solving Absolute Value Problems

The absolute value of a number is its distance from zero on the number line.For the equation $$|x|=5$$, we saw that both 5 and $$−5$$ are five units from zero on the number line. $\begin & & \ & & \ \nonumber \end$ What about the inequality $$|x|\leq 5$$?For the equation |x|=5,|x|=5, we are looking for all numbers that make this a true statement.

The absolute value of a number is its distance from zero on the number line.For the equation $$|x|=5$$, we saw that both 5 and $$−5$$ are five units from zero on the number line. $\begin & & \ & & \ \nonumber \end$ What about the inequality $$|x|\leq 5$$?

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Example $$\Page Index$$ The ideal diameter of a rod needed for a machine is 80 mm.

The actual diameter can vary from the ideal diameter by 0.009 mm.

The absolute value of a number is its distance from zero on the number line. We saw that the numbers whose distance is less than or equal to five from zero on the number line were $$−5$$ and 5 and all the numbers between $$−5$$ and 5 (Figure $$\Page Index$$).

Now we want to look at the inequality $$|x|\geq 5$$.

We just saw that both 5 and −5−5 are five units from zero on the number line. $$\begin & \\ & \\ \end$$ The solution can be simplified to a single statement by writing $$x=\pm 5$$. If two algebraic expressions are equal in absolute value, then they are either equal to each other or negatives of each other.

This is read, “ To solve an absolute value equation, we first isolate the absolute value expression using the same procedures we used to solve linear equations. The property for absolute value equations says that for any algebraic expression, Let’s look now at what happens when we have an absolute value inequality.

He now plans to buy new equipment that guarantees the thickness of the tortilla within 0.005 inches.

If the ideal thickness of the tortilla is 0.04 inches, what thickness of tortillas will be guaranteed?

Example $$\Page Index$$ The ideal diameter of a rod needed for a machine is 75 mm.

The actual diameter can vary from the ideal diameter by 0.05 mm.

## Comments Solving Absolute Value Problems

• ###### Solving absolute value problems -

There are basically 2 types of absolute value problems We'll call the first one a "lesser-than" problem and the second one a "greater-than" problem. In both types of problems, there are 2 inequalities you have to solve, giving you 2 solution sets for x. The bottom line is the final solution set for x has to contain both solution sets for x.…

• ###### SparkNotes Absolute Value Problems

Problems. Problem What is the value of + if = - 2 ? Problem Find the solution set of 3 + 4 = 24 from the replacement set { -12, -4, 0, 4, 12}. Problem Find the solution set of = 3 from the replacement set { -6, -5, -1, 0, 3}. Problem Find the solution set of 4 - 2 = - 12 from the replacement set { -1, 0, 1, 3, 5}. Be Book-Smarter.…

• ###### Working with Absolute Value Independent Practice Worksheet

Name _____ Date _____ Tons of Free Math Worksheets at © with Absolute Value - Independent Practice Worksheet…

• ###### Algebra precalculus - How do I solve a double absolute value.

Answers. With an inequality where there are 2 absolute value equations on either side of the inequality i.e. All you have to do is make both positive, and then a separate inequality where 1 absolute value is negative. Both Positive The 's cancel out, so this equation is no use to us.…

• ###### How to Solve Absolute Value Equations With a Number on the.

Solve an absolute value equation that contains a number outside the absolute value bars by algebraically moving that number to the side of the equation opposite the variable. Eliminate the absolute value by creating two equations from the expression, representing the positive and negative possibilities for the terms within the bars.…

• ###### Solving Absolute Value Equations - Kuta Software LLC

Solving Absolute Value Equations Date_____ Period____ Solve each equation. 1 3 x = 9 2 −3r = 9 3 b 5 = 1 4 −6m = 30 5 n 3 = 2 6 −4 + 5x = 16 7 −2r − 1 = 11 8 1 − 5a = 29 9 −2n + 6 = 6 10 v + 8 − 5 = 2-1-…