Question 66·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

A restriction is a value the original equation can't use. Factor the first denominator as a difference of squares:

The denominators are zero at $x=4$ or $x=-4$, so neither value can be a solution.

Type $y=\frac{6x^2}{x^2-16}-\frac{3x}{x-4}$ and $y=\frac{2}{x+4}$ on separate lines. Click their crossing, about $(0.667,0.429)$. The $x$-coordinate, $0.667$, is the input that makes the original sides equal; $0.429$ is their shared output.

Of the choices, $\frac23\approx0.667$ matches the crossing's positive $x$-coordinate. It isn't either forbidden value, so $x=\frac23$ satisfies the original equation. Choice B.

The least common denominator is $(x-4)(x+4)$: it contains every original denominator's factors. Multiply every term by it. This is reversible only for the allowed values $x\ne4,-4$:

Factor $3x$ out of both terms on the left, keeping the whole $x+4$ in parentheses:

Subtract both terms inside the parentheses:

Now the same factor appears on both sides.

You already ruled out $x=4$, so $x-4$ isn't zero for any possible solution. Divide both sides by $x-4$:

Clearing fractions can make a forbidden value look like a solution; keep the original restrictions when you cancel.

Divide by $3$:

This value doesn't make any original denominator zero, so it solves the original equation. Choice B.