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Consider the system of inequalities đť‘¦ is less than the absolute value of đť‘Ą plus four and đť‘¦ is greater than or equal to the absolute value of negative two đť‘Ą plus two.
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Which of the shaded areas A, B, C, or D represents the solution to the system?
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Here are our two inequalities.
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We have a few things to consider.
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One, the graph itself will have a certain look about it.
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An absolute value graph looks like a V, which we can see in A, B, C and D.
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So we need to decide which graph is correct.
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And notice, some lines are solid and some of the lines are dashed.
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If the inequality has a less than or greater than symbol, it needs to be a dash line, which we have here.
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If we have a less than or equal to or greater than or equal to, there should be a solid line, which we have here.
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Next, letâ€™s begin plugging in numbers to decide which graph goes with which inequality.
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So we will plug in the values of negative one, zero, and one for đť‘Ą.
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The absolute value of negative one plus four will be equal to the absolute value of three, because negative one plus four is three.
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Then the absolute value signs make any number positive, and the absolute value of three will be three.
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The absolute value of zero plus four will be equal to the absolute value of four, which is equal to four.
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And then lastly, the absolute value of one plus four would be equal to the absolute value of five, which is equal to five.
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So at negative one, we should be at three.
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At zero, we should be at four.
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And at one, we should be at five.
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And this should be our dashed graph.
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So letâ€™s look at option A.
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At negative one, we are at three.
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At zero, weâ€™re at four.
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And at one, weâ€™re at five.
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This does not match the dashed graph.
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Now letâ€™s look at option B.
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Option B works.
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At negative one, weâ€™re at three.
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At zero, weâ€™re at four.
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And at one for đť‘¦, we are at five.
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And this is indeed with a dash line.
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It also works for option C, but does not work for option D.
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So we can already eliminate option A and option D.
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So now letâ€™s begin looking at our other inequality.
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So if we plug in a negative one, we have negative two times negative one plus two, and then we take the absolute value.
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So negative two times negative one will be positive two, and two plus two is four, and the absolute value of four is equal to four.
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Next, we have negative two times zero plus two.
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Negative two times zero is zero, and zero plus two is two, and the absolute value of two is two.
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Now we have the absolute value of negative two times one plus two, and negative two times one is negative two, and negative two plus two is zero, and the absolute value of zero is zero.
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At negative one for đť‘Ą, we should be at four for đť‘¦.
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At zero for đť‘Ą, we should be at two for đť‘¦.
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And lastly, at one, we should be at zero.
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And this should be the solid lines.
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So looking at option B, at negative one, weâ€™re at four.
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At zero, weâ€™re at two.
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And at one, weâ€™re at zero.
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So this works.
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Letâ€™s check option C.
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At negative one, weâ€™re at four, which works.
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At zero, weâ€™re at two.
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But at one, weâ€™re not at zero.
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So option C does not work.
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So right now, we know that option B is our answer.
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However, what about the shading part?
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For shading, a less than or less than or equal to sign means we shade below that line.
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For a greater than or greater than or equal to sign, we shade above that line.
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So for the dash line, we need to shade below it.
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Or we could think underneath it.
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And then for the solid line, we need to shade above that line.
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And wherever these shadings would intersect would be our answer, which would indeed be here.
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Therefore, shaded area B represents the solution to this system.