Question

Solving 1/x^2 + 1/y^2 - 13=0 and 1/x + 1/y - 5=0, find the roots​

Answer 1

Answer:

1/2 & 1/3

Step-by-step explanation:

Solving 1/x^2 + 1/y^2 - 13=0 and 1/x + 1/y - 5=0

1/x² + 1/y² - 13 = 0

=> 1/x² + 1/y² = 13

1/x + 1/y - 5 = 0

=> 1/x + 1/y = 5

Squaring both sides

1/x² + 1/y² + 2/xy = 25

=> 13 + 2/xy = 25

=> 2/xy = 12

1/x² + 1/y² = 13

= (1/x - 1/y)² + 2/xy  = 13

=> (1/x - 1/y)² = 13 - 2/xy

=> (1/x - 1/y)² = 13 - 12

=> (1/x - 1/y)² = 1

=>  (1/x - 1/y) = ±1

1/x + 1/y = 5  & (1/x - 1/y) = 1

=> Adding both 2/x = 6  => x = 1/3   & y = 1/2

1/x + 1/y = 5  & (1/x - 1/y) = -1

=>  Adding both 2/x = 4  => x = 1/2  &   y = 1/3

so roots are   1/2 & 1/3