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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 $8$.
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