Solve for x (x^2-16)/(x-4)=x+4

$\frac{x^{2} - 16}{x - 4} = x + 4$

Factor each term.

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Rewrite $16$ as $4^{2}$ .

$\frac{x^{2} - 4^{2}}{x - 4} = x + 4$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

$\frac{(x + 4) (x - 4)}{x - 4} = x + 4$

Reduce the expression by cancelling the common factors.

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Reduce the expression $\frac{(x + 4) (x - 4)}{x - 4}$ by cancelling the common factors.

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Cancel the common factor.

$\frac{(x + 4) (x - 4)}{x - 4} = x + 4$

Rewrite the expression.

$\frac{x + 4}{1} = x + 4$

Divide $x + 4$ by $1$ .

$x + 4 = x + 4$

Move all terms containing $x$ to the left side of the equation.

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Subtract $x$ from both sides of the equation.

$x + 4 - x = 4$

Combine the opposite terms in $x + 4 - x$ .

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Subtract $x$ from $x$ .

$0 + 4 = 4$

Add $0$ and $4$ .

$4 = 4$

Since $4 = 4$ , the equation will always be true for any value of $x$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for x (x^2-16)/(x-4)=x+4

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