Question

If 3x+4y=16 and 3x-4y=4,find the value of xy. ((Hint:Use the formula (p+q)^2-(p-q)^2=4pq))..

Answer 1

Step-by-step explanation:

★ᏀᏆᏙᎬΝ:-

$\sf 3x+4y=16$

$\sf 3x-4y=4$

★Ꭲᝪ ᖴᏆᑎᗞ:-

$\sf xy$

★ʜɪɴᴛ:-

$\sf (p+q)^2 - (p-q)^2 = 4pq$

★ន០ɭ⩏Ƭɨ០⩎:-

$\sf 3x + 4y = 16$

Squaring both the sides of the equation will give,

$\sf (3x+4y)^2=(16)^2$

$\sf (3x+4y)^2 = 256$

__________________________

$\tt 3x - 4y = 4$

Squaring both the sides,

$\tt (3x-4y)^2 = (4)^2$

$\tt (3x-4y)^2 = 16$

___________________________

Let p = 3x

q = 4y

Using the hint,

$\sf (p+q)^2 - (p-q)^2 = 4pq$

$\sf (3x + 4y)^2 - (3x-4y)^2 = 4 \times 3x \times 4y$

$\sf (16)^2 - (4)^2 = 48xy$

$\sf 256 - 16 = 48xy$

$\sf 240 = 48xy$

$\sf xy = \dfrac{240}{48}$

$\sf xy = 5$

Value of xy is 5

Answer 2

Answer:let 3x be p and 4y be q

3x+4y^2-3x-4y^2=43x4y

9x^2+16y^2-9x^2-16y^2=4 3x4y

-12xy=4

xy=4/-12

xy=-3

Step-by-step explanation: