If $\frac{x}{2 y = 3}$ $x/(2y=3)$ , what is the value of $x^{2} - 36 y^{2}$ $x^2-36y^2$ ?

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If $\frac{x}{2 y = 3}$ $x/(2y=3)$ , what is the value of $x^{2} - 36 y^{2}$ $x^2-36y^2$ ?

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Answer 1 · 2017-07-22T02:49:18

I kind of do not get if the equation you want is $\frac{x}{2 y} = 3$ $x/(2y)=3$ , but just feel free to remind me if I misread your question.

Explanation:

The important thing to do is to use cross-multiply to help the values in the first equation; since there is a constant (number) equal to it.

$x = 6 y$ $x=6y$

Move it over to make it into a monomial.

$x - 6 y = 0$ $x-6y=0$

And since this the negative conjugate to the polynomial equation given, we can infer that the equation is equal to 0; according to the zero property of multiplication.

$x^{2} - 36 y^{2} = (x + 6 y) (x - 6 y)$ $x^2-36y^2= (x+6y)(x-6y)$

$(x - 6 y) = 0;$ $(x-6y)=0;$

$x^{2} - 36 y^{2} = 0$ $x^2-36y^2=0$

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If $\frac{x}{2 y = 3}$ $x/(2y=3)$ , what is the value of $x^{2} - 36 y^{2}$ $x^2-36y^2$ ?

1 Answer

Explanation:

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If $\frac{x}{2 y = 3}$ $x/(2y=3)$ , what is the value of $x^{2} - 36 y^{2}$ $x^2-36y^2$ ?