x, y are natural numbers such that x > y. Also x + y + xy = 80, then value of x is
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Answered by
76
well, there are so many solutions when we assume x and y belongs to all real numbers.
but when x and y are natural number such that x > y , then we get only one specific solution.
also you should know this question can be solved with hit and trial method.
given, x + y + xy = 80
Let x + y = n then, xy = 80 - n
now, (n - x)x = 80 - n
nx - x² = 80 - n
x² - nx + 80 - n = 0
D = n² - 4(80 - n) ≥ 0
n² - 320 + 4n ≥ 0
n² + 4n - 320 ≥ 0
(n - 16)(n + 20) ≥ 0
n ≥ 16 => x + y ≥ 16
and 80 - n ≤ 64 => xy ≤ 64
we have condition : x + y ≥ 16 & xy ≤ 64
take two random numbers such that x > y and also it follows x + y + xy = 80
if we take x = 26 and y = 2
then, we see x + y = 28 ≥ 16
xy = 26 × 2 = 52 ≤ 64
and x + y + xy = 26 + 2 + 26 × 2 = 80
hence, x = 26 and y = 2
so, value of x = 26.
but when x and y are natural number such that x > y , then we get only one specific solution.
also you should know this question can be solved with hit and trial method.
given, x + y + xy = 80
Let x + y = n then, xy = 80 - n
now, (n - x)x = 80 - n
nx - x² = 80 - n
x² - nx + 80 - n = 0
D = n² - 4(80 - n) ≥ 0
n² - 320 + 4n ≥ 0
n² + 4n - 320 ≥ 0
(n - 16)(n + 20) ≥ 0
n ≥ 16 => x + y ≥ 16
and 80 - n ≤ 64 => xy ≤ 64
we have condition : x + y ≥ 16 & xy ≤ 64
take two random numbers such that x > y and also it follows x + y + xy = 80
if we take x = 26 and y = 2
then, we see x + y = 28 ≥ 16
xy = 26 × 2 = 52 ≤ 64
and x + y + xy = 26 + 2 + 26 × 2 = 80
hence, x = 26 and y = 2
so, value of x = 26.
Answered by
15
Given:
x and y are natural numbers.
Natural numbers: 1,2,3,...
Condition:
- x>y
- x+y+xy=80
Deductions:
If x=2, then y<2.
In order to satisfy the second condition,
Consider,
x=26
y=2
Condition 1 is satisfied because 26<1
26+2+(26*2)=80
28+52=80
Hence, Condition 2 is also satisfied.
Therefore,
x=26 and y=2.
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