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Question
Solve the following pair of equations:
`x/a + y/b = a + b, x/a^2 + y/b^2 = 2, a, b ≠ 0`
Solution
Given pair of linear equations is
`x/a + y/b` = a + b .....(i)
And `x/a^2 + y/b^2` = 2, a, b ≠ 0 ......(ii)
On multiplying equation (i) by `1/a` and then subtracting from equation (ii), we get
`x/a^2 + y/b^2 = 2`
`x/a^2 + y/(ab) = 1 + b/a`
– – –
`y(1/b^2 - 1/(ab)) = 2 - 1 - b/a`
⇒ `y((a - b)/(ab^2)) = 1 - b/a = ((a - b)/a)`
⇒ y = `(ab^2)/a`
⇒ y = b2
Now, put the value of y in equation (ii), we get
`x/a^2 + b^2/b^2` = 2
⇒ `x/a^2` = 2 – 1 = 1
⇒ x = a2
Hence, the required values of x and y are a2 and b2, respectively.
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