If $ x - y = 1 $ and $ x^2 + y^2 = 41 $, then the value of $ x + y $ will be

### Concept & Formula This problem links the sum and difference of two variables via their squared terms. You need to use the expansion formulas for both $ (x - y)^2 $ and $ (x + y)^2 $ to bridge the gap using the product $ xy $. The key identities are: $$ (x - y)^2 = x^2 + y^2 - 2xy $$ $$ (x + y)^2 = x^2 + y^2 + 2xy $$ ### Step-by-Step Solution **Step 1: Find the value of $ 2xy $** We are given $ x - y = 1 $ and $ x^2 + y^2 = 41 $. Use the squared difference formula: $$ (x - y)^2 = x^2 + y^2 - 2xy $$ Substitute the given values into the formula: $$ (1)^2 = 41 - 2xy $$ $$ 1 = 41 - 2xy $$ Rearrange to solve for $ 2xy $: $$ 2xy = 41 - 1 $$ $$ 2xy = 40 $$ **Step 2: Find the value of $ (x + y)^2 $** Now, use the squared sum formula: $$ (x + y)^2 = x^2 + y^2 + 2xy $$ Substitute the known value of $ x^2 + y^2 $ (which is 41) and our calculated value for $ 2xy $ (which is 40): $$ (x + y)^2 = 41 + 40 $$ $$ (x + y)^2 = 81 $$ **Step 3: Solve for $ x + y $** Take the square root of both sides. Remember that squaring either a positive or a negative number yields a positive result. $$ x + y = \pm \sqrt{81} $$ $$ x + y = \pm 9 $$ ### Exam Strategy & Shortcut For small integer squares, look for Pythagorean triplets or familiar square sums. We need two numbers whose squares add to 41. Testing small squares: $ 1, 4, 9, 16, 25, 36 $. Notice that $ 16 + 25 = 41 $. These are the squares of 4 and 5. Check the first condition: $ 5 - 4 = 1 $. This matches! So, $ x = 5 $ and $ y = 4 $. Their sum is $ 5 + 4 = 9 $. Since variables can also be negative (e.g., $ -4 $ and $ -5 $ where $ -4 - (-5) = 1 $), the sum can also be $ -9 $. Thus, $ \pm 9 $. ### Common Pitfall The most common trap is calculating $ (x + y)^2 = 81 $ and simply selecting 9 as the answer, forgetting that the square root of 81 is algebraically $ \pm 9 $. Exam setters often provide both options to catch this exact oversight. ### Final Answer **Therefore, the correct answer is $ \pm 9 $.**