Question

x/a + y/b = a+b ; x/a^2 + y/b^2 = 2ab , b is not equal to 0

Answer 1

Answer :

Given,

x/a + y/b = a + b ...(i)

x/a² + y/b² = 2ab ...(ii)

Now, multiplying (i) by 1/a and (ii) by 1, we get

x/a² + y/ab = (a + b)/a

x/a² + y/b² = 2ab

On subtraction, we get

y/ab - y/b² = (a + b)/a - 2ab

➽ (1/ab - 1/b²)y = (a + b - 2a²b)/a

➽ {(b - a)/ab²} y = (a + b - 2a²b)/a

➽ {(b - a)/b²} y = (a + b - 2a²b)

➽ y = b²(a + b - 2a²b)/(b - a)

Now, putting y = b²(a + b - 2a²b)/(b - a) in (i), we get

x/a + b²(a + b - 2a²b)/(b - a) = a + b

➽ x/a = (a + b) - b²(a + b - 2a²b)/(b - a)

➽ x/a = (ab + b² - a² - ab - ab² - b³ + 2a²b³)/(b - a)

➽ x/a = (b² - a² - ab² - b³ + 2a²b³)/(b - a)

➽ x = a(b² - a² - ab² - b³ + 2a²b³)/(b - a)

∴ The required solution be

x = a(b² - a² - ab² - b³ + 2a²b³)/(b - a)

y = b²(a + b - 2a²b)/(b - a)

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