Question

Solve the following pair of equations:

`x/a + y/b = a + b, x/a^2 + y/b^2 = 2, a, b ≠ 0`

Answer 1

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.