Question

hlo plz help me fasttttttt .its urgent

Answer 1

$\underline\bold{\huge{SOLUTION \: :}}$



4) $\boxed{FIRST\:\:CASE :}$

m (2/3)² + n (2/3) + 6 = 0

=> 4m/9 + 2n/3 + 6 = 0

=> (4m + 6n + 54)/9 = 0

=> 4m + 6n + 54 = 0 .......... (i)

$\boxed{SECOND\:\:CASE :}$

m (1)² + n(1) + 6 = 0

=> m + n + 6 = 0 .............. (ii)


Now, let equation (i) be multiplied by 1 and equation (ii) be multiplied by 4. Subtracting them, we get :

4m+6n+54 - (4m+4n+24) = 0 - 0

=> 4m + 6n + 54 - 4m - 4n - 24 = 0

=> 2n + 30 = 0

=> n = -15

Putting n = -15 in equation (ii), we get :

m - 15 + 6 = 0

=> m - 9 = 0

=> m = 9

Hence, the answer is : m = 9 and n = -15
______________________________

5) $\boxed{FIRST\:\:CASE :}$

a (3)² + b(3) - 9 = 0

=> 9a + 3b - 9 = 0

=> 3 (3a + b - 3) = 0

=> 3a + b - 3 = 0 .......... (i)


$\boxed{SECOND\:\:CASE :}$

a (-3)² + b(-3) - 9 = 0

=> 9a - 3b - 9 = 0

=> 3 (3a - b - 3) = 0

=> 3a - b - 3 = 0 .......... (ii)


Subtracting both the equations, we get :

3a+b-3 - (3a-b-3) = 0 - 0

=> 3a+b-3-3a+b+3 = 0

=> 2b = 0

=> b = 0

Putting b = 0 in equation (ii), we get :

3a - b - 3 = 0

=> 3a - 0 - 3 = 0

=> 3 a - 3 = 0

=> 3 a = 3

=> a = 1


Hence, the answer is : a = 1 and b = 0