About Absolute Value Inequalities Part 2:
In some cases, we may have to solve more advanced absolute value inequalities. When solving absolute value inequalities with two absolute value operations involved, we will set the arguments of the absolute value operations equal to zero. We will then solve these equations and use the solutions to split up our number line into three intervals. From there, we will replace our absolute value operation with the expression itself if the argument is 0 or positive in the interval or the negative of the expression if the argument is negative in the interval. We will then go through each possible inequality and check to see if the solution lies in our given interval. Once we have verified all the correct solutions, we can report our answer. Lastly, we may see an absolute value inequality with a rational expression involved as the argument of the absolute value operation.
Test Objectives
- Demonstrate the ability to solve an absolute value inequality with two absolute value operations
- Demonstrate the ability to solve an absolute value inequality with rational expressions
#1:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}|x - 3| > |2x - 5|$$
$$b)\hspace{.2em}|3x - 4| > |x - 2|$$
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#2:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}x + 3 ≥ |2x - 3| + 1$$
$$b)\hspace{.2em}|5x + 12| > |x - 9| + 1$$
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#3:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}|3x - 5| - |2x - 7| ≥ 6$$
$$b)\hspace{.2em}||2x - 1| + 3| ≤ 4$$
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#4:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}||x - 2| - 3| ≤ 2$$
$$b)\hspace{.2em}|x^2 + 3x - 4| ≤ 6$$
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#5:
Instructions: solve each inequality, write in interval notation, graph.
$$a)\hspace{.2em}\left|\frac{5x - 2}{x - 1}\right| > 1$$
$$b)\hspace{.2em}\frac{|2x - 9|}{|x + 3|}≤ 1$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}2 < x < \frac{8}{3}$$ $$\left(2, \frac{8}{3}\right)$$
$$b)\hspace{.2em}x < 1 \hspace{.2em}\text{or}\hspace{.2em}x > \frac{3}{2}$$ $$(-\infty, 1) ∪ \left(\frac{3}{2}, \infty \right)$$
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#2:
Solutions:
$$a)\hspace{.2em}\frac{1}{3}≤ x ≤ 5$$ $$\left[\frac{1}{3}, 5\right]$$
$$b)\hspace{.2em}x < -\frac{11}{2}\hspace{.2em}\text{or}\hspace{.2em}x > - \frac{1}{3}$$ $$\left(-\infty, -\frac{11}{2}\right) ∪ \left(-\frac{1}{3}, \infty\right)$$
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#3:
Solutions:
$$a)\hspace{.2em}x ≤ -8 \hspace{.2em}\text{or}\hspace{.2em}x ≥ 4$$ $$(-\infty, -8] ∪ [4, \infty)$$
$$b)\hspace{.2em}0≤x≤1$$$$[0, 1]$$
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#4:
Solutions:
$$a)\hspace{.2em}{-}3 ≤ x ≤ 1$$$$ \text{or}$$$$ 3 ≤ x ≤ 7$$ $$[-3, 1] ∪ [3, 7]$$
$$b)\hspace{.2em}{-}5 ≤ x ≤ -2$$ $$\text{or}$$ $${-}1 ≤ x ≤ 2$$ $$[-5, -2] ∪ [-1, 2]$$
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#5:
Solutions:
$$a)\hspace{.2em}x < \frac{1}{4}\hspace{.2em}\text{or}\hspace{.2em}\frac{1}{2}< x < 1 \hspace{.2em}\text{or}\hspace{.2em}x > 1$$ $$\left(-\infty, \frac{1}{4}\right) ∪ \left(\frac{1}{2}, 1\right) ∪ (1, \infty)$$
$$b)\hspace{.2em}2 ≤ x ≤ 12$$ $$[2, 12]$$
