x/a + y/b=a+b………!
X/a^2+y/b^2=2
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There is a typing mistake in the given question.
Given a /x + b/y = a+b --(1) and a/x² + b/y² = 2 ---(2)
To solve for x and y.
Let 1/x = X and 1/y = Y.
Then a X + b Y = a+b ---(3)
=> X = (a+b - bY)/a ---(4)
a X² + b Y² = 2 ---(5)
Substitute X from (4) in (5) and simplify.
a [(a+b) - bY]²/a² + b Y² = 2
b(b+a) Y² - 2 b(a+b)Y + (a+b)²-2a = 0
Y = [ b(a+b) + √ {b(a+b) × {b(a+b) - (b+a)² + 2a } } ]/ { b(b+a)}
= 1 + √{a(2-a-b)} / √{b(a+b) }
y = 1/Y = √b(a+b) / [ √(ab+b²) + √(a (2-a-b)) ]
Substituting this value in (1) we get:
x = √(ab(a+b)) / [ √(ab(a+b)) -+ √(a(2-a-b))
Given a /x + b/y = a+b --(1) and a/x² + b/y² = 2 ---(2)
To solve for x and y.
Let 1/x = X and 1/y = Y.
Then a X + b Y = a+b ---(3)
=> X = (a+b - bY)/a ---(4)
a X² + b Y² = 2 ---(5)
Substitute X from (4) in (5) and simplify.
a [(a+b) - bY]²/a² + b Y² = 2
b(b+a) Y² - 2 b(a+b)Y + (a+b)²-2a = 0
Y = [ b(a+b) + √ {b(a+b) × {b(a+b) - (b+a)² + 2a } } ]/ { b(b+a)}
= 1 + √{a(2-a-b)} / √{b(a+b) }
y = 1/Y = √b(a+b) / [ √(ab+b²) + √(a (2-a-b)) ]
Substituting this value in (1) we get:
x = √(ab(a+b)) / [ √(ab(a+b)) -+ √(a(2-a-b))
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