Question

7. If 16x^2 +25y^2 = 401 , xy=1. Find 4x + 5y. ​

Answer 1

Answer:

value of 4x + 5y is 441

Step-by-step explanation:

given that :

$16x^2 + 25 y^2 = 401$

also given that :

xy = 1

we have to find the value of  4x + 5y

let's square the problem;

(4x + 5y)²

⇒ (4x)² + (5y)² + 2(4x)(5y)

⇒ 16x² + 25y² + 40xy

substitute value of $(16x^2 + 25y^2) = 401$  in above equation:

⇒ 16x² + 25y² + 40xy

⇒ 401 + 40xy

substitute value of  'xy=1' in above equation;

⇒ 401 + (40)(1)

⇒ 401 + 40

⇒ 441

value of 4x + 5y is 441