solving inequalities with absolute value


When we take the opposite of a quantity, we must be careful with the signs and to add parentheses where needed.

Where are the numbers whose distance from zero is greater than or equal to five?

You may want to work through the tutorials on solving equations with absolute value before solving the questions below. Solve for x. \\ { |u| \geq a } &{\quad \text{is equivalent to}} &{ u\leq −a \quad \text{ or } \quad u\geq a} The absolute value of a number n is written as \(|n|\) and \(|n|\geq 0\) for all numbers. That is, learn the rules and apply it correctly. [latex]\left|2x+3\right|+9\leq 7[/latex].

&{} &{} &{} \\ \end{array}\). We can generalize this to the following property for absolute value equations. \\ \end{array}\). x < 4. Again we will look at our definition of absolute value. Now we want to look at the inequality \(|x|\geq 5\). Interval notation: [latex]\left(-\infty, -3\right)\cup\left(3,\infty\right)[/latex]. \(\begin{array} {ll} {} &{|\frac{2}{3}x−4|=−8} \\ {\text{Isolate the absolute value term.}} For example, \(|x−3|=|2x+1|\). \(\begin{array} {ll} {} &{|y|=−6} \\ {} &{\text{No solution}} \\ \end{array}\) In the following video you will see an example of solving multi-step absolute value inequalities involving an and situation. If two algebraic expressions are equal in absolute value, then they are either equal to each other or negatives of each other. Some of our absolute value equations could be of the form \(|u|=|v|\) where u and v are algebraic expressions. \(\begin{array} {ll} {} &{} &{|5x−1|=|2x+3|} &{} \\ {} &{} &{} &{} \\ {\text{Write the equivalent equations.}} Graph the solution and write the solution in interval notation: SOLVE ABSOLUTE VALUE INEQUALITIES WITH \(<\) OR \(\leq\), \[\begin{array} {lll} {|u|2\). &{} &{} &{} \\ {\text{We leave the check to you.}} Remember that if we end up with an absolute value greater than or less than a negative number, there is no solution. 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 \(u=a\). But [latex]5[/latex] and [latex]−5[/latex] would work and so would all of the values extending to the left of [latex]−3[/latex] and to the right of [latex]3[/latex]. [latex]\begin{array}{r}\underline{-18}<\underline{2y}<\underline{\,6\,}\\2\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,2\,\,\\-9<\,\,y\,\,\,\,<\,3\end{array}[/latex], Inequality notation: [latex] \displaystyle -9<\,\,y\,\,<3[/latex], Interval notation: [latex]\left(-9,3\right)[/latex].

Inequalities containing absolute value can be solved by rewriting them using compound inequalities. \\ {\text{and so}} &{u=v \text{ or } u = −v} &{\text{or}} &{u=−v \text{ or } u = −(−v)} [latex]{x}\le\text{−a}[/latex] or [latex]{x}\ge{ a}[/latex], [latex]\left(-\infty,-a\right]\cup\left[a,\infty\right)[/latex], [latex]\left| x \right|\gt\text{a}[/latex], [latex]\displaystyle{x}\lt\text{−a}[/latex] or [latex]{x}\gt{ a}[/latex], [latex]\left(-\infty,-a\right)\cup\left(a,\infty\right)[/latex], Express solutions to inequalities containing absolute value, Identify solutions for absolute inequalities where there are no solutions. a less than) is very different from solving an inequality with a > (i.e. Once we isolate the absolute value expression we rewrite it as the two equivalent equations.

If we subtract 3 from both sides, we get: x + 3 − 3 < 7 − 3 . The graph of the solution divides the number line into three sections. \\ \end{array}\). Check the solutions in the original equation to be sure they work.
The absolute value of a quantity can never be a negative number, so there is no solution to the inequality. Again both \(−5\) and 5 are five units from zero and so are included in the solution. What range of diameters will be acceptable to the customer without causing the rod to be rejected? If the difference from the specifications exceeds the tolerance, the item is rejected. To verify, check a value in each section of the number line showing the solution. See Figure. Graph the solution and write the solution in interval notation: \(|x|<1\). There are four cases involved when solving absolute value inequalities. What range of diameters will be acceptable to the customer without causing the rod to be rejected?

Since an absolute value is always positive, there are no solutions to this equation. Solve absolute value inequalities with “less than”, Solve absolute value inequalities with “greater than”. Step 2: If there is only one absolute value in the inequality, solve it in both areas (where the argument of the absolute value is negative, and where it is non-negative). \[\begin{array} {lll} Again, you could think of the number line and what values of x are greater than [latex]3[/latex] units away from zero.

The solution can be simplified to a single statement by writing \(x=\pm 5\). For any positive value of a and x, a single variable, or any algebraic expression: Let us look at a few more examples of inequalities containing absolute value. Remember that an absolute value cannot be a negative number. The steps for solving an absolute value equation are summarized here. Constructive Media, LLC.

We use first party cookies on our website to enhance your browsing experience, and third party cookies to provide advertising that may be of interest to you. Graph the solution and write the solution in interval notation. Solve \(|3x−4|\geq 2\). { |u| >a } &{\quad \text{is equivalent to}} &{ u<−a \quad \text{ or } \quad u>a} \[|\text{actual-ideal}|\leq \text{tolerance} \nonumber\]. [latex]\left|x+3\right|\gt4[/latex].

Sum and product of the roots of a quadratic equations Algebraic identities. By … This leads us to the following property for equations with two absolute values. In the following video, you will see an example of solving multi-step absolute value inequalities involving an or situation. SOLVE ABSOLUTE VALUE INEQUALITIES WITH \(>\) OR \(\geq\).

You can accept or reject cookies on our website by clicking one of the buttons below. Ex 2: Solve and Graph Absolute Value inequalities Mathispower4u . Adding or Subtracting a Value. Solve \(|4x−3|\geq 5\). The actual diameter can vary from the ideal diameter by \(0.075\) mm. All the numbers between \(−5\) and 5 are less than five units from zero (Figure \(\PageIndex{2}\)). To verify, check a value in each section of the number line showing the solution. Solving Absolute Value Inequalties with Less Than, Solving Absolute Value Inequalties with Greater Than. [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.

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