How to Solve ABS(5x-2) > OR = 23 for X. (An Algebra Problem for Student Nurses)
In order to solve an Absolute Value Inequality, the solution is a set of numbers that is infinite. In order to understand what the solution set looks like, it is best to put it the form of a graph. This article will show how to do this.
Instructions
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The first step we need to do to solve the inequality, ABS(5x-2)> or = (>=)23, is to rewrite the inequality, by removing the Absolute Value Bars. To do this, we can use the Rule of; [ABS(x)>=K, then (x<= -K OR x>= K)]. When we rewrite the inequality, it will be [(5x-2)<= -23 OR (5x-2)>= 23]. Please click on the image for a better understanding.
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We will now Solve for X, using the two new inequalities. The first step is to Add 2 to both sides of the inequality sign, for both inequalities, and then combine Like Terms; [5x-2+2>=23+2 OR 5x-2+2<=-23+2] = [5x>=25 OR 5x<=-21]. This will put the Constant Term on the right side of the inequalities, and the X term on the left side. Now, we will divide both sides of the two inequalities by 5; [5x/5>=25/5 OR 5x/5<=-21/5]. This will give us our Final Answer, which is [x>=5 OR x<=-21/5]. Please click on the image for a better understanding.
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Now we need to write the Solution, in Interval Notation, and then Graph it. The Solution is (x>=5 OR x<=-21/5). The Solution in Interval Notation is (-infinity, -21/5]U[5,infinity). To graph the solution, we need to draw the number line, and graph the points, -21/5, which is the same as -4.2, and the point 5 on it. Draw a closed circle( a circle in which the interior is shaded), over -21/5 with an arrow pointing to the left, and a closed in circle over 5 with an arrow pointing to the right. Please click on the image for a better understanding.
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Tips & Warnings
A Closed Circle on the Initial point of an arrow indicates that that point is included in the solution set. If the circle is not closed then it is an open circle, indicating that, that point is NOT included in the solution set.
