Question

√3x+✓2y=2√2,√2x-√3y=3√3 then values of x and y are​

Answer 1

Answer:

Basic Algebra Identities

(a + b)2 = a2 + b2 + 2ab

(a – b)2 = a2 + b2 – 2ab

a2 – b2 = (a + b)(a – b)

a2 + b2 = (a + b)2 – 2ab = (a – b)2 + 2ab

a3 + b3 = (a + b)(a2 – ab + b2)

a3 – b3 = (a – b)(a2 + ab + b2)

(a + b)3 = a3 + 3ab(a + b) + b3

(a – b)3 = a3 – 3ab(a – b) – b3

Answer 2

Answer:

$\sqrt{3x + } \sqrt2y = 2 \sqrt{2}$

$\sqrt{2 x} - \sqrt{3y} = 3 \sqrt{3}$

Taking the square root of Equation 1 and Equation 2

$3x + 2y = 2 \times 2 \\ \\ 2x - 3y = 3 \times 3 \\ \\ 3x + 2y = 4 \\ 2x - 3y = 9$

Multiplying by 2 in equation 1 and multiplying by 3 in equation 2

$6x + 4y = 8 \\ 6y - 9y = 27$

- dividing by

$13y = 19$

Y=19/13

Considering y in Equation 1

$3x + 2y = 4$

3x + 2 × 19/13= 4

3x+38/13=4

3x = 4/1 - 38/13

3x = 52-38/13

3x = 14/13

x= 14/39

X = 14/39

Y = 19/13

Answer .