{"id":57756,"date":"2025-08-28T17:11:11","date_gmt":"2025-08-28T11:41:11","guid":{"rendered":"https:\/\/e-gmat.com\/blogs\/?p=57756"},"modified":"2025-09-04T19:46:26","modified_gmt":"2025-09-04T14:16:26","slug":"what-values-of-x-have-a-corresponding-value-of-y-that-satisfies-both","status":"publish","type":"post","link":"https:\/\/e-gmat.com\/blogs\/what-values-of-x-have-a-corresponding-value-of-y-that-satisfies-both\/","title":{"rendered":"What values of x have a corresponding value of y that satisfies both xy > 0 and xy ="},"content":{"rendered":"A <\/span> 4<\/span> min read <\/span><\/span>\n

Question:<\/strong> What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ?<\/p>\n\n\n\n

Options:<\/strong> <\/p>\n\n\n\n

A. x <= 1 <\/p>\n\n\n\n

B. -1 < x <= 0 <\/p>\n\n\n\n

C. 0 < x <= 1 <\/p>\n\n\n\n

D. x > 1 <\/p>\n\n\n\n

E. All real numbers<\/p>\n\n\n\n

Solution Section<\/h2>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

  • Translate the problem requirements: We need values of x where there exists a corresponding y such that both xy > 0 (x and y have the same sign) AND xy = x + y (the product equals the sum)<\/li>
  • Solve for y in terms of x: Use the equation xy = x + y to express y as a function of x, determining when this relationship is valid<\/li>
  • Apply the sign constraint: Use the condition xy > 0 to determine which values of x allow for valid y values that satisfy both conditions<\/li>
  • Verify the solution set: Test boundary values and confirm the range of x that works<\/li><\/ul>\n\n\n\n

    Execution of Strategic Approach<\/h3>\n\n\n\n
    1. Translate the problem requirements<\/li><\/ol>\n\n\n\n

      Let’s break down what this problem is asking in plain English. We need to find values of x where we can find a corresponding y such that BOTH of these things happen: \u2022 xy > 0: This means x and y have the same sign (both positive or both negative) \u2022 xy = x + y: The product of x and y equals their sum<\/p>\n\n\n\n

      Think of it this way: imagine you have two numbers x and y. When you multiply them together, you get the same result as when you add them together. And on top of that, both numbers must have the same sign.<\/p>\n\n\n\n

      Process Skill: TRANSLATE – Converting the mathematical conditions into clear requirements we can work with<\/p>\n\n\n\n

      1. Solve for y in terms of x<\/li><\/ol>\n\n\n\n

        Now let’s use the equation xy = x + y to find y in terms of x. This will help us understand when such a y can exist.<\/p>\n\n\n\n

        Starting with: xy = x + y<\/p>\n\n\n\n

        Let’s rearrange this to get all terms with y on one side: xy – y = x<\/p>\n\n\n\n

        Factor out y: y(x – 1) = x<\/p>\n\n\n\n

        So: y = x\/(x – 1)<\/p>\n\n\n\n

        But wait! This formula only works when x \u2260 1, because we can’t divide by zero when x – 1 = 0.<\/p>\n\n\n\n

        Let’s check what happens when x = 1: If x = 1, our original equation xy = x + y becomes 1 \u00b7 y = 1 + y, which gives us y = 1 + y. This is impossible since it would mean 0 = 1.<\/p>\n\n\n\n

        So for any valid solution, we need x \u2260 1, and y = x\/(x – 1).<\/p>\n\n\n\n

        Process Skill: MANIPULATE – Algebraic rearrangement to express y in terms of x<\/p>\n\n\n\n

        1. Apply the sign constraint<\/li><\/ol>\n\n\n\n

          Now we use the condition xy > 0, which means x and y must have the same sign.<\/p>\n\n\n\n

          We know y = x\/(x – 1), so xy = x \u00b7 x\/(x – 1) = x\u00b2\/(x – 1)<\/p>\n\n\n\n

          For xy > 0, we need x\u00b2\/(x – 1) > 0.<\/p>\n\n\n\n

          Since x\u00b2 is always positive (except when x = 0), the sign of this expression depends entirely on the sign of (x – 1): \u2022 If x – 1 > 0 (meaning x > 1), then x\u00b2\/(x – 1) > 0 \u2713 \u2022 If x – 1 < 0 (meaning x < 1), then x\u00b2\/(x – 1) < 0 \u2717<\/p>\n\n\n\n

          But wait, we also need to check what happens when x = 0: If x = 0, then y = 0\/(0-1) = 0, and xy = 0 \u00b7 0 = 0, which doesn’t satisfy xy > 0.<\/p>\n\n\n\n

          Therefore, we need x > 1 for both conditions to be satisfied.<\/p>\n\n\n\n

          Process Skill: APPLY CONSTRAINTS – Using the sign requirement to determine valid x values<\/p>\n\n\n\n

          1. Verify the solution set<\/li><\/ol>\n\n\n\n

            Let’s test our conclusion with a concrete example:<\/p>\n\n\n\n

            Try x = 2: Then y = 2\/(2-1) = 2\/1 = 2 Check: xy = 2 \u00b7 2 = 4, and x + y = 2 + 2 = 4 \u2713 Also: xy = 4 > 0 \u2713<\/p>\n\n\n\n

            Try x = 0.5: Then y = 0.5\/(0.5-1) = 0.5\/(-0.5) = -1 Check: xy = 0.5 \u00b7 (-1) = -0.5, and x + y = 0.5 + (-1) = -0.5 \u2713 But: xy = -0.5 < 0 \u2717<\/p>\n\n\n\n

            This confirms that x > 1 is indeed the correct condition.<\/p>\n\n\n\n

            1. Final Answer<\/li><\/ol>\n\n\n\n

              The values of x that satisfy both conditions are x > 1.<\/p>\n\n\n\n

              Looking at the answer choices, this corresponds to choice D. x > 1.<\/p>\n\n\n\n

              Common Faltering Points<\/h3>\n\n\n\n

              Errors while devising the approach<\/strong><\/p>\n\n\n\n

              1. Misinterpreting the constraint xy > 0<\/strong> Students often misunderstand what xy > 0 means. They might think it simply means “x and y are both positive” instead of understanding that it means “x and y have the same sign” (both positive OR both negative). This leads them to only consider positive values and miss part of the analysis.<\/li>
              2. Treating the conditions as separate instead of simultaneous<\/strong> Some students approach this by first finding all x where xy > 0 could be satisfied, then separately finding where xy = x + y could be satisfied, rather than recognizing that we need to find x values where BOTH conditions can be satisfied by the SAME y value simultaneously.<\/li>
              3. Forgetting to check for domain restrictions<\/strong> Students may dive straight into algebraic manipulation without considering whether certain values of x might make the problem undefined or impossible to solve, particularly when x = 1 makes the denominator zero.<\/li><\/ol>\n\n\n\n

                Errors while executing the approach<\/strong><\/p>\n\n\n\n

                1. Algebraic manipulation errors when solving for y<\/strong> When rearranging xy = x + y to get y = x\/(x-1), students commonly make errors such as incorrectly factoring out y or making sign errors. They might get y = x\/(1-x) instead of y = x\/(x-1).<\/li>
                2. Incorrect sign analysis for the inequality<\/strong> When analyzing x\u00b2\/(x-1) > 0, students often forget that x\u00b2 is always non-negative, and incorrectly conclude that both the numerator and denominator need to be analyzed for sign changes. They may create incorrect sign charts or miss the fact that x = 0 makes the entire expression equal to 0.<\/li>
                3. Missing the special case x = 0<\/strong> Even after correctly finding that x > 1 works, students often forget to explicitly check what happens when x = 0, since x\u00b2 = 0 makes the analysis different from other cases where x\u00b2 > 0.<\/li><\/ol>\n\n\n\n

                  Errors while selecting the answer<\/strong><\/p>\n\n\n\n

                  1. Including the boundary point x = 1<\/strong> After determining that we need x – 1 > 0, some students incorrectly conclude that x \u2265 1 satisfies the conditions, forgetting that x = 1 was specifically excluded because it makes the original equation impossible to solve.<\/li>
                  2. Confusing which inequality direction to use<\/strong> Students might correctly identify that the boundary is at x = 1 but then select x < 1 instead of x > 1, especially if they made a sign error earlier and didn’t catch it through verification.<\/li>
                  3. <\/li><\/ol>\n","protected":false},"excerpt":{"rendered":"

                    Question: What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ? Options: A. x <= 1 B. -1 < x <= 0 C. 0 < x <= 1 D. x > 1 E. All real numbers Solution Section Solution: Translate the problem […]<\/p>\n","protected":false},"author":102457,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_et_pb_use_builder":"off","_et_pb_old_content":"","_et_gb_content_width":"","ub_ctt_via":""},"categories":[44,100],"tags":[],"featured_image_src":null,"author_info":{"display_name":"Kashish Garg","author_link":"https:\/\/e-gmat.com\/blogs\/author\/kashish\/"},"yoast_head":"\nWhat values of x have a corresponding value of y that satisfies both xy > 0 and xy =<\/title>\n<meta name=\"description\" content=\"What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ? 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