Question

Melanie used cross multiplication to correctly solve a rational equation. She found one valid solution and one extraneous solution. If 3 is the extraneous solution, which equation could she have solved?

A. The equation is 4/x-3 = x/10 because 3 makes a denominator equal zero and is not a solution of the equation derived from cross multiplying.


B. The equation is x-3/4 = 2x-6/4x because 3 makes a numerator equal zero and is a solution if the equation derived from cross multiplying.


C. The equation is 8/x^2-9 = 5/2x-6 because 3 makes the denominator equal zero and is a solution of the equation derived from cross multiplying.


D. The equation is 8/x+3 = 12/4x-3 because 3 is a solution of both the original equation and the equation derived from cross multiplying.

Answer 1

The equation is 4/x-3 = x/10 because 3 makes a denominator equal zero and is not a solution of the equation derived from cross multiplying. ( A )

Explanation:

The equation is 4/x-3 = x/10 because 3 makes a denominator equal zero and is not a solution of the equation derived from cross multiplying.

An extraneous solution is a solution that is derived from an equation that is not the original equation and when its value is plugged back into the original equation it isn't a root of the equation