Refer to this question below Thank You.
1 answer:
Let's solve this problem using substitution. Given that x-y=8, x = 8 + y. (Then x^2 = 64 + 16y + y^2) This other equation is (x-2)^2 + (y-1)^2 = 25. Easier to substitute 8 + y for x in (x-2)^2: (8 + y - 2)^2 + (y-1)^2 = 25 (6 + y)^2 + (y-1)^2 = 25 36 + 12y + y^2 + y^2 - 2y + 1 = 25 Re-writing this in descending powers of y: 2y^2 + 10y + 36 + 1 = 25 Then 2y^2 + 10y - 12 = 0 Reduce by division by 2: y^2 + 5y - 6 = 0 = (y+3)(y+2) = 0 Then y=-3 and y=-2. From each of these we get x: x = 8 + y So x = 8 - 3 = 5 and x = 8 - 2 = 6. There are common solutions. Try (5, -3) and (6, -2). Do these points satisfy both of the given equations? If they do, you've shown that we have common solutions.
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