If #x/(2y=3)#, what is the value of #x^2-36y^2#?

1 Answer
Peter L.
Jul 22, 2017

I kind of do not get if the equation you want is #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=6y#

Move it over to make it into a monomial.

#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^2-36y^2= (x+6y)(x-6y)#

#(x-6y)=0;#

#x^2-36y^2=0#

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