If $2x + 3y = 14$ and $2x - 3y = 2$, find value of $xy$. [Hint: Use $(2x+3y)^2 - (2x-3y)^2 = 24xy$]

Given:

$2x + 3y = 14$ and $2x - 3y = 2$

To do:

We have to find the value of $xy$.

Solution:

The given expressions are $2x + 3y = 14$ and $2x - 3y = 2$. Here, we have to find the value of $xy$. So, by squaring and subtracting using the identities $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$, we can find the required value.

$(a+b)^2=a^2+2ab+b^2$.............(I)

$(a-b)^2=a^2-2ab+b^2$.............(II)

Let us consider,

$2x + 3y = 14$

Squaring on both sides, we get,

$(2x + 3y)^2 = (14)^2$

$(2x)^2+2(2x)(3y)+(3y)^2=196$            [Using (I)]

$4x^2+12xy+9y^2=196$..........(III)

Now,

$2x - 3y = 2$

Squaring on both sides, we get,

$(2x - 3y)^2 = (2)^2$

$(2x)^2-2(2x)(3y)+(3y)^2=4$            [Using (II)]

$4x^2-12xy+9y^2=4$..........(IV)

Subtracting (IV) from (III), we get,

$(4x^2+12xy+9y^2)-(4x^2-12xy+9y^2)=196-4$

$4x^2-4x^2+12xy+12xy+9y^2-9y^2=192$

$24xy=192$

$xy=\frac{192}{24}$

$xy=8$

Hence, the value of $xy$ is $8$.