Question

Hey There ! Kindly answer the 10th one . Thank q

Answer 1

Question :

If 3x + 2y = 12 and xy = 6 , find the value of $9{x}^{2}$ + $4{y}^{2}$

Answer :

= 3x + 2y = 12

• Squaring on both sides

= ${(3x+2y)}^{2}$ = 144

= $9{x}^{2}$ + 2×3x×2y+ $4{y}^{2}$ =144

= $9{x}^{2}$ + 12xy $4{y}^{2}$ = 144

= $9{x}^{2}$ + $4{y}^{2}$= 144 - 12(xy)

= $9{x}^{2}$ + $4{y}^{2}$ = 144 - 12(6)

= $9{x}^{2}$ + $4{y}^{2}$ = 144 - 72

= $9{x}^{2}$ + $4{y}^{2}$ = 72

Thus, the required answer is 72 respectively.

___________________________

Answer 2

Step-by-step explanation:

3x + 2y = 12

Squaring on both sides

= {(3x+2y)}^{2}(3x+2y)

2

= 144

= 9{x}^{2}9x

2

+ 2×3x×2y+ 4{y}^{2}4y

2

=144

= 9{x}^{2}9x

2

+ 12xy 4{y}^{2}4y

2

= 144

= 9{x}^{2}9x

2

+ 4{y}^{2}4y

2

= 144 - 12(xy)

= 9{x}^{2}9x

2

+ 4{y}^{2}4y

2

= 144 - 12(6)

= 9{x}^{2}9x

2

+ 4{y}^{2}4y

2

= 144 - 72

= 9{x}^{2}9x

2

+ 4{y}^{2}4y

2

= 72