Question

5. if x and y are two positive real numbers, such
that as x ² +9y2 = 369 & xy = 60 then, find
the value of x- 3y​

Answer 1

Answer:

please refer to the attachment

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Answer 2

Given,

$x^{2} + 9y^{2} = 369\\xy = 60$

To Find,

$x - 3y =?$

Solution,

We can solve the question using the following steps:

$x^{2} + 9y^{2} = 369\\x^{2} + (3y)^{2} = 369$

Now,

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

             $= x^{2} + 9y^{2} - 6xy$

Substituting the given values,

$(x - 3y)^{2} = 369 - 6*60$

               $= 369 - 360$

                $= 9$

Now,

$x - 3y = \sqrt{9} =$ ±$3$

Hence, the value of x - 3y is equal to ±3.