x and y are natural numbers such that y ≤ 6. If x1 and x2 be two possible values of x for which x²y + x + y is divisible by xy² + y + 7; then find the value of (x1 + x2).
Answers
Given : x and y are natural numbers such that y ≤ 6 x1 and x2 be two possible values of x for which x²y + x + y is divisible by xy² + y + 7;
To find : value of (x1 + x2).
Solution:
x²y + x + y is divisible by xy² + y + 7
y ≤ 6
y = 1
x²y + x + y = x² + x + 1
xy² + y + 7 = x + 1 + 7 = x + 8
not Divisible
( will be divisible if x = - 8 will result x² + x + 1 = 0)
but here 57
y = 2
x²y + x + y = 2x² + x + 2
xy² + y + 7 = 4x + 9
Similarly not Divisible
y = 3
x²y + x + y = 3x² + x + 3
xy² + y + 7 = 9x + 10
Similarly not Divisible
y = 4
x²y + x + y = 4x² + x + 4
xy² + y + 7 = 16x + 11
Similarly not Divisible
y = 5
x²y + x + y = 5x² + x + 5
xy² + y + 7 = 25x + 12
Similarly not Divisible
y = 6
x²y + x + y = 6x² + x + 6
xy² + y + 7 = 36x + 13
Similarly not Divisible
There seems some mistake in data
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