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If a ,b, x, y are natural numbers such that a2 + b2 = 25, x2 + y2 = 13, then (ax + by) + (ay + bx) has exactly
  • a)
    8 values
  • b)
    4 values
  • c)
    2 values
  • d)
    1 value
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
If a ,b, x, y are natural numbers such that a2 + b2 = 25, x2 + y2 = 1...
Explanation:

Given:
a^2 + b^2 = 25
x^2 + y^2 = 13

Finding the value of (ax + by) + (ay + bx):
Let's consider the expression (ax + by) + (ay + bx):
(ax + by) + (ay + bx) = a(x + y) + b(y + x) [Rearranging terms]
= a(x + y) + b(x + y) [Rearranging terms again]
= (a + b)(x + y) [Factoring out common terms]
Now, we know that a^2 + b^2 = 25 and x^2 + y^2 = 13. Since a, b, x, y are natural numbers, the only possible values for a, b, x, y are 1, 2, 3, 4, 5.
Substitute these values into a^2 + b^2 = 25:
- If a = 3 and b = 4, (a + b) can be 7 and (x + y) can be 2 or 5. Thus, the expression can have only 1 value: 7 * 2 = 14 or 7 * 5 = 35.
- If a = 4 and b = 3, (a + b) can be 7 and (x + y) can be 2 or 5. Thus, the expression can have only 1 value: 7 * 2 = 14 or 7 * 5 = 35.
Therefore, the expression (ax + by) + (ay + bx) has exactly 1 value. Hence, the correct answer is option 'D'.
Free Test
Community Answer
If a ,b, x, y are natural numbers such that a2 + b2 = 25, x2 + y2 = 1...
It is given that a2 + b2 = 25 and a & b are natural numbers
Only natural number solution to the above equation is either a = 4, b = 3 or a=3, b=4
Similarly for the equation x2 + y2 = 13, only natural number solution is either x=2, y=3 or x=3,y=2
We can observe that (a+b) will be always 7 and (x+y) will be 5
(ax + by) + (ay + bx) = (a+b)(x+ y) = 7 x 5 = 35. => Only 1 value
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If a ,b, x, y are natural numbers such that a2 + b2 = 25, x2 + y2 = 13, then (ax + by) + (ay + bx) has exactlya)8 valuesb)4 valuesc)2 valuesd)1 valueCorrect answer is option 'D'. Can you explain this answer?
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