Question

calculate the values of 2. f * x ^ 2 + y ^ 2 = 5/8 * and * 4/3 * xy = 1/4 (a) (x + y) (b) (x-y) (c) (x^2-y^2)​

Answer 1

(a) 1

(b) 1/2

(c) 1/2

Explanation:

`x^2 + y^2 = 5/8 (1)\\(4)/(3)xy = (1)/(4) (2)\\\\\textrm {From (2) by multiplying both sides by 3/4 we get}\\\\xy = (3)/(16) (3)\\\textrm{We know that } (x + y)^2 = x^2 + 2xy + y^2 \\\textrm{Adding 2xy to the LHS and RHS of equation 1 gisues us }\\x^2 + y^2 + 2xy = (5)/(8) + 2 * (3)/(16) = (5)/(8) + (3)/(8) = (8)/(8) = 1`

`(x + y)^2 = 1 == > x+y = √(1) = 1\\\\(x - y)^2 = x^2 -2xy + y^2\\\\\textrm{Subtracting 2xy from both sides of Equation 1}\\x^2 + y^2 - 2xy = (5)/(8) - 2*(3)/(16) = (5)/(8) - (3)/(8) = (2)/(8) = (1)/(4) \\x-y = \sqrt{(1)/(4)} = (1)/(2)\\\\x^2 - y^2 = (x+y)(x-y) = 1 . (1)/(2) = (1)/(2)`