Ex 4: Solve and Graph Absolute Value inequalities (Requires Isolating Abs. SOLVE ABSOLUTE VALUE INEQUALITIES WITH \(>\) OR \(\geq\). Write the equivalent compound inequality. What about the inequality \(|x|\leq 5\)? For example: Figure \(\PageIndex{1}\) illustrates this idea. \(\begin{array} {ll} {} &{} &{|5x−1|=|2x+3|} &{} \\ {} &{} &{} &{} \\ {\text{Write the equivalent equations.}} Graph the solution and write the solution in interval notation. \\ { |u| >a } &{\quad \text{is equivalent to}} &{ u<−a \quad \text{ or } \quad u>a} \\ { |u| \geq a } &{\quad \text{is equivalent to}} &{ u\leq −a \quad \text{ or } \quad u\geq a}
The graph of this inequality will have two closed circles, at [latex]4[/latex] and [latex]−4[/latex]. In the following video, you will see examples of how to solve and express the solution to absolute value inequalities involving both and and or. The first step to solving absolute inequalities is to isolate the absolute value. x < 4. &{|\text{actual-ideal}|\leq \text{tolerance}} \\ {} &{|x−60|\leq 0.075} \\ {\text{Rewrite as a compound inequality.}} [latex]\begin{array}{r}\underline{3\left| 2y+6 \right|}\,<\underline{36}\\3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\\\,\,\,\,\,\,\,\,\,\left| 2y+6 \right|<12\end{array}[/latex]. If the number on the other side of the inequality sign is negative, your equation either has no solution or all real numbers as solutions. There are four cases involved when solving absolute value inequalities. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Graph the solution and write the solution in interval notation: \(|x|<9\). This leads us to the following property for equations with two absolute values. As with equations, there may be instances where there is no solution to an inequality. &{−0.075\leq x−60\leq 0.075} \\ {\text{Solve the inequality.}} Fill in \(<,>,<,>,\) or \(=\) for each of the following pairs of numbers. \\ \end{array}\). All the numbers between \(−5\) and 5 are less than five units from zero (Figure \(\PageIndex{2}\)). Let us apply what you know about solving equations that contain absolute value and what you know about inequalities to solve inequalities that contain absolute value.
What range of diameters will be acceptable to the customer without causing the rod to be rejected? Have questions or comments? Again we will look at our definition of absolute value. Again we will look at our definition of absolute value. Choose numbers such as −8,−8, 1, and 9. In the following video you will see an example of solving multi-step absolute value inequalities involving an and situation. In this final section of the Solving chapter we will solve inequalities that involve absolute value. Check the solutions in the original equation to be sure they work. Adopted or used LibreTexts for your course? Inequalities containing absolute value can be solved by rewriting them using compound inequalities. By the end of this section, you will be able to: Before you get started, take this readiness quiz. The actual diameter can vary from the ideal diameter by 0.009 mm. Where are the numbers whose distance from zero is greater than or equal to five? Prealgebra solving inequalities lessons with lots of worked examples and practice problems.
Where are the numbers whose distance is less than or equal to 5?
Usually there is a certain tolerance of the difference from the specifications that is allowed.
Remember that if we end up with an absolute value greater than or less than a negative number, there is no solution. We can generalize this to the following property for absolute value equations. [latex]3[/latex] and [latex]−3[/latex] are also solutions because each of these values is less than [latex]4[/latex] units away from [latex]0[/latex]. This inequality is read, “the absolute value of x is less than or equal to [latex]4[/latex].” If you are asked to solve for x, you want to find out what values of x are [latex]4[/latex] units or less away from [latex]0[/latex] on a number line. ABSOLUTE VALUE INEQUALITIES WITH \(>\) OR \(\geq\), \[\begin{array} {lll} {\text{if}} &{\quad |u|>a,} &{\quad \text{then } u<−a \text{ or } u>a} \\ {\text{if}} &{\quad |u|\geq a,} &{\quad \text{then } u\leq −a \text{ or } u\geq a} \\ \nonumber \end{array}\]. \\ {\text{then}} &{|u|=v} &{\text{or}} &{|u|=−v}
\(\begin{array} {ll} {} &{|z|=0} \\ {\text{Write the equivalent equations.}} How would we solve them? \[\begin{array} {ll} {\text{if}} &{|u|=|v|} \\ {\text{then}} &{u=−v\text{ or }u=v} \\ \nonumber \end{array}\]. To understand more about how we and our advertising partners use cookies or to change your preference and browser settings, please see our Global Privacy Policy.
Divide by [latex]2[/latex] to isolate the variable. Solve absolute value inequalities with “less than”, Solve absolute value inequalities with “greater than”.
Solving quadratic equations by quadratic formula. Solve Absolute Value Inequalities with “Less Than” Let’s look now at what happens when we have an absolute value inequality. Write the solution using interval notation.
In the last video that follows, you will see an example of solving an absolute value inequality where you need to isolate the absolute value first. After solving an inequality, it is often helpful to check some points to see if the solution makes sense. \\ {\text{and so}} &{u=v \text{ or } u = −v} &{\text{or}} &{u=−v \text{ or } u = −(−v)} { |u| >a } &{\quad \text{is equivalent to}} &{ u<−a \quad \text{ or } \quad u>a} Once we isolate the absolute value expression we rewrite it as the two equivalent equations. The absolute value of a number is its distance from zero on the number line. As we will see the process for solving inequalities with a < (i.e. Again both \(−5\) and 5 are five units from zero and so are included in the solution. [latex] \displaystyle x+3<-4\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,x+3>4[/latex], [latex]\begin{array}{r}x+3<-4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x+3>4\\\underline{\,\,\,\,-3\,\,\,\,\,-3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,-3\,\,-3}\\x\,\,\,\,\,\,\,\,\,<-7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,>1\\\\x<-7\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x>1\,\,\,\,\,\,\,\end{array}[/latex]. &{\text{The diameter of the rod can be between}} \\ {} &{59.925 mm \text{ and } 60.075 mm.} &{} &{} &{} \\ \end{array}\). Graph the solution and write the solution in interval notation. Divide both sides by 3 to isolate the absolute value. [latex] \displaystyle \begin{array}{r}\,\,\,\,\,\left| x+3 \right|>4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| x+3 \right|>4\\\left| -10+3 \right|>4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| 5+3 \right|>4\\\,\,\,\,\,\,\,\,\,\,\left| -7 \right|>4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| 8 \right|>4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,7>4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8>4\end{array}[/latex], Inequality notation: [latex] \displaystyle x<-7\,\,\,\,\,\text{or}\,\,\,\,\,x>1[/latex], Interval notation: [latex]\left(-\infty, -7\right)\cup\left(1,\infty\right)[/latex], Solve for y. The steps for solving an absolute value equation are summarized here. &{59.925\leq x\leq 60.075} \\ {\text{Answer the question.}} An item must be made with near perfect specifications. Solve \(|x|>1\). 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 \(\PageIndex{4}\)). Graph the solution and write the solution in interval notation. \(5\) is 5 units away from 0, so \(|5|=5\). Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. \(\begin{array} {ll} {} &{|x|=8} \\ {\text{Write the equivalent equations.}} What range of diameters will be acceptable to the customer without causing the rod to be rejected? They are the solutions to the equation. By … The diameter of the rod can be between 79.991 and 80.009 mm.
Graphing absolute value equations Combining like terms Inequalities containing absolute value can be solved by rewriting them using compound inequalities. &{7x−1=−3} &{} &{3x=4} \\ {} &{7x=−2} &{} &{x=43} \\ {} &{x=−27} &{\text{or}} &{x=43} \\ {\text{Check.}} And that is our solution: x < 4. \\ \nonumber \end{array}\]. Again, you could think of the number line and what values of x are greater than [latex]3[/latex] units away from zero. We know \(−5\) and 5 are both five units from zero. When we take the opposite of a quantity, we must be careful with the signs and to add parentheses where needed. So are [latex]1[/latex] and [latex]−1[/latex],[latex]0.5[/latex] and [latex]−0.5[/latex], and so on—there are an infinite number of values for x that will satisfy this inequality. CAUTION: In all cases, the assumption is that the … Absolute Value Inequalities Read More » Solving quadratic equations by completing square.
The next step is to decide whether you are working with an or inequality or an and inequality.
If the difference from the specifications exceeds the tolerance, the item is rejected. Let us start with a simple inequality. &{\text{No solution}} \\ \end{array}\). {\text{if}} &{|u|=|v|} &{} &{} The property for absolute value equations says that for any algebraic expression, u, and a positive real number, a, if \(|u|=a\), then \(u=−a\) or \(u=a\). Write the absolute value inequality using the “less than” rule. The situation is a little different when the inequality sign is “greater than” or “greater than or equal to.” Consider the simple inequality [latex]\left|x\right|>3[/latex]. [latex] \displaystyle \begin{array}{r}\,\,\,\left| x+3 \right|>4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| x+3 \right|>4\\\left| -7+3 \right|=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| 1+3 \right|=4\\\,\,\,\,\,\,\,\left| -4 \right|=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| 4 \right|=4\\\,\,\,\,\,\,\,\,\,\,\,\,4=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4=4\end{array}[/latex]. Nature of the roots of a quadratic equations. [latex]4[/latex] and [latex]−4[/latex] are both four units away from [latex]0[/latex], so they are solutions. For the equation \(|x|=5\), we saw that both 5 and \(−5\) are five units from zero on the number line. [latex] \displaystyle \begin{array}{r}3\left| 2y+6 \right|-9<27\\\underline{\,\,+9\,\,\,+9}\\3\left| 2y+6 \right|\,\,\,\,\,\,\,\,<36\end{array}[/latex]. Absolute Value Equations And Inequalities. \\ \end{array}\). \(\begin{array} {ll} {\text{If}} &{|x|=5} \\ {\text{then}} &{x=−5\text{ or }x=5} \\ \end{array}\). We are looking for the numbers whose distance from zero is 5. Everything we’ve learned about solving inequalities still holds, but we must consider how the absolute value impacts our work. If the inequality is less than a number, we will use and.
If the inequality is greater than a number, we will use or. &{z=−0\text{ or }z=0} \\ {\text{Since }−0=0,} &{z=0} \\ \end{array}\) Remember, an absolute value is always positive! Sum and product of the roots of a quadratic equations Algebraic identities. Solve for x. Graph the solution and write the solution in interval notation: SOLVE ABSOLUTE VALUE INEQUALITIES WITH \(<\) OR \(\leq\), \[\begin{array} {lll} {|u|
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