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Master gmat 2010 part 57
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The correct answer is (D). To answer the question, there’s no way around doing some pencil-work. You need to solve for x in each equation, then square it. Let’s start with statement (2), which is a bit easier to work with: 2x 1 5 5 1 2x 5 24 x 5 22 x2 5 4 Now let’s tackle statement (1), which presents a more complex equation: ~x 2 3!2 5 ~x 1 7!2 ~x 2 3!~x 23! 5 ~x 1 7!~x 1 7! x2 2 6x 1 9 5 x2 1 14x 1 49 20x 5 240 x 5.