Find the number of integral solutions of \( \frac{1}{x} + \frac{3}{y} = \frac{1}{29} \) with the condition \( x > 0 \).
Answer (2)
To find integral solutions for x 1 + y 3 = 29 1 with 0"> x > 0 , we reformulated the equation and determined that for positive integers, multiple values for k can generate solutions for x and y . Each valid integer k leads to a unique pair of ( x , y ) fulfilling the conditions.
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To find the number of integral solutions for the equation:
x 1 + y 3 = 29 1
with the condition that 0"> x > 0 , we first need to manipulate the equation to eliminate the fractions. Let's rewrite the equation:
Multiplying throughout by 29 x y to clear the denominators, we have:
29 y + 87 x = x y .
Rearranging gives:
x y − 87 x − 29 y = 0.
We can try solving this equation by using Simon's Favorite Factoring Trick. We add and subtract the same quantity to complete the factorization:
x y − 87 x − 29 y + 2523 = 2523.
Now, factor by grouping:
( x − 29 ) ( y − 87 ) = 2523.
So, we need to find the pairs ( a , b ) such that ( x − 29 ) = a and ( y − 87 ) = b where a ⋅ b = 2523 .
The number 2523 factors into:
2523 = 3 2 × 281.
Since 2523 is a composite number, we determine the positive pairs ( a , b ) such that their product equals 2523. The factors are:
( 1 , 2523 )
( 3 , 841 )
( 9 , 281 )
( 281 , 9 )
( 841 , 3 )
( 2523 , 1 )
Each pair ( a , b ) provides a value of x and y given by:
x = a + 29 y = b + 87
Ensuring 0"> x > 0 doesn’t provide further restrictions since all found a + 29 are positive.
Therefore, there are 6 integral solutions ( x , y ) for the given equation x 1 + y 3 = 29 1 with 0"> x > 0 . These correspond to the 6 pairs of factors.
