Solve for x x/(x+4)+5/(4-x)=1/(x-4)

$\frac{x}{x + 4} + \frac{5}{4 - x} = \frac{1}{x - 4}$

Step 1

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x + 4, 4 - x, x - 4$

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

The factor for $x + 4$ is $x + 4$ itself.

$(x + 4) = x + 4$

$(x + 4)$ occurs $1$ time.

The factor for $4 - x$ is $4 - x$ itself.

$(4 - x) = 4 - x$

$(4 - x)$ occurs $1$ time.

The factor for $x - 4$ is $x - 4$ itself.

$(x - 4) = x - 4$

$(x - 4)$ occurs $1$ time.

The LCM of $x + 4, 4 - x, x - 4$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x + 4) (4 - x) (x - 4)$

Step 2

Multiply each term in $\frac{x}{x + 4} + \frac{5}{4 - x} = \frac{1}{x - 4}$ by $(x + 4) (4 - x) (x - 4)$ to eliminate the fractions.

Multiply each term in $\frac{x}{x + 4} + \frac{5}{4 - x} = \frac{1}{x - 4}$ by $(x + 4) (4 - x) (x - 4)$ .

$\frac{x}{x + 4} ((x + 4) (4 - x) (x - 4)) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Simplify the left side.

Simplify each term.

Cancel the common factor of $x + 4$ .

Factor $x + 4$ out of $(x + 4) (4 - x) (x - 4)$ .

$\frac{x}{x + 4} ((x + 4) ((4 - x) (x - 4))) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Cancel the common factor.

$\frac{x}{x + 4} ((x + 4) ((4 - x) (x - 4))) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Rewrite the expression.

$x ((4 - x) (x - 4)) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Rewrite $4$ as $- 1 (- 4)$ .

$x ((- 1 (- 4) - x) (x - 4)) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Factor $- 1$ out of $- x$ .

$x ((- 1 (- 4) - (x)) (x - 4)) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Factor $- 1$ out of $- 1 (- 4) - (x)$ .

$x (- 1 (- 4 + x) (x - 4)) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Reorder terms.

$x (- 1 (x - 4) (x - 4)) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Raise $x - 4$ to the power of $1$ .

$x (- 1 ({(x - 4)}^{1} (x - 4))) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Raise $x - 4$ to the power of $1$ .

$x (- 1 ({(x - 4)}^{1} {(x - 4)}^{1})) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x (- 1 {(x - 4)}^{1 + 1}) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Add $1$ and $1$ .

$x (- 1 {(x - 4)}^{2}) + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Rewrite using the commutative property of multiplication.

$- x {(x - 4)}^{2} + \frac{5}{4 - x} ((x + 4) (4 - x) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Cancel the common factor of $4 - x$ .

Factor $4 - x$ out of $(x + 4) (4 - x) (x - 4)$ .

$- x {(x - 4)}^{2} + \frac{5}{4 - x} ((4 - x) ((x + 4) (x - 4))) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Cancel the common factor.

$- x {(x - 4)}^{2} + \frac{5}{4 - x} ((4 - x) ((x + 4) (x - 4))) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Rewrite the expression.

$- x {(x - 4)}^{2} + 5 ((x + 4) (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Expand $(x + 4) (x - 4)$ using the FOIL Method.

Apply the distributive property.

$- x {(x - 4)}^{2} + 5 (x (x - 4) + 4 (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Apply the distributive property.

$- x {(x - 4)}^{2} + 5 (x \cdot x + x \cdot - 4 + 4 (x - 4)) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Apply the distributive property.

$- x {(x - 4)}^{2} + 5 (x \cdot x + x \cdot - 4 + 4 x + 4 \cdot - 4) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

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

Reorder the factors in the terms $x \cdot - 4$ and $4 x$ .

$- x {(x - 4)}^{2} + 5 (x \cdot x - 4 x + 4 x + 4 \cdot - 4) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Add $- 4 x$ and $4 x$ .

$- x {(x - 4)}^{2} + 5 (x \cdot x + 0 + 4 \cdot - 4) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Add $x \cdot x$ and $0$ .

$- x {(x - 4)}^{2} + 5 (x \cdot x + 4 \cdot - 4) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Simplify each term.

Multiply $x$ by $x$ .

$- x {(x - 4)}^{2} + 5 (x^{2} + 4 \cdot - 4) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Multiply $4$ by $- 4$ .

$- x {(x - 4)}^{2} + 5 (x^{2} - 16) = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Apply the distributive property.

$- x {(x - 4)}^{2} + 5 x^{2} + 5 \cdot - 16 = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Multiply $5$ by $- 16$ .

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = \frac{1}{x - 4} ((x + 4) (4 - x) (x - 4))$

Simplify the right side.

Cancel the common factor of $x - 4$ .

Factor $x - 4$ out of $(x + 4) (4 - x) (x - 4)$ .

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = \frac{1}{x - 4} ((x - 4) ((x + 4) (4 - x)))$

Cancel the common factor.

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = \frac{1}{x - 4} ((x - 4) ((x + 4) (4 - x)))$

Rewrite the expression.

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = (x + 4) (4 - x)$

Expand $(x + 4) (4 - x)$ using the FOIL Method.

Apply the distributive property.

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = x (4 - x) + 4 (4 - x)$

Apply the distributive property.

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = x \cdot 4 + x (- x) + 4 (4 - x)$

Apply the distributive property.

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = x \cdot 4 + x (- x) + 4 \cdot 4 + 4 (- x)$

Simplify and combine like terms.

Simplify each term.

Move $4$ to the left of $x$ .

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = 4 \cdot x + x (- x) + 4 \cdot 4 + 4 (- x)$

Rewrite using the commutative property of multiplication.

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = 4 x - x \cdot x + 4 \cdot 4 + 4 (- x)$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = 4 x - (x \cdot x) + 4 \cdot 4 + 4 (- x)$

Multiply $x$ by $x$ .

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = 4 x - x^{2} + 4 \cdot 4 + 4 (- x)$

Multiply $4$ by $4$ .

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = 4 x - x^{2} + 16 + 4 (- x)$

Multiply $- 1$ by $4$ .

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = 4 x - x^{2} + 16 - 4 x$

Subtract $4 x$ from $4 x$ .

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = - x^{2} + 16 + 0$

Add $- x^{2}$ and $0$ .

$- x {(x - 4)}^{2} + 5 x^{2} - 80 = - x^{2} + 16$

Step 3

Solve the equation.

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

Add $x^{2}$ to both sides of the equation.

$- x {(x - 4)}^{2} + 5 x^{2} - 80 + x^{2} = 16$

Simplify each term.

Rewrite ${(x - 4)}^{2}$ as $(x - 4) (x - 4)$ .

$- x ((x - 4) (x - 4)) + 5 x^{2} - 80 + x^{2} = 16$

Expand $(x - 4) (x - 4)$ using the FOIL Method.

Apply the distributive property.

$- x (x (x - 4) - 4 (x - 4)) + 5 x^{2} - 80 + x^{2} = 16$

Apply the distributive property.

$- x (x \cdot x + x \cdot - 4 - 4 (x - 4)) + 5 x^{2} - 80 + x^{2} = 16$

Apply the distributive property.

$- x (x \cdot x + x \cdot - 4 - 4 x - 4 \cdot - 4) + 5 x^{2} - 80 + x^{2} = 16$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$- x (x^{2} + x \cdot - 4 - 4 x - 4 \cdot - 4) + 5 x^{2} - 80 + x^{2} = 16$

Move $- 4$ to the left of $x$ .

$- x (x^{2} - 4 \cdot x - 4 x - 4 \cdot - 4) + 5 x^{2} - 80 + x^{2} = 16$

Multiply $- 4$ by $- 4$ .

$- x (x^{2} - 4 x - 4 x + 16) + 5 x^{2} - 80 + x^{2} = 16$

Subtract $4 x$ from $- 4 x$ .

$- x (x^{2} - 8 x + 16) + 5 x^{2} - 80 + x^{2} = 16$

Apply the distributive property.

$- x \cdot x^{2} - x (- 8 x) - x \cdot 16 + 5 x^{2} - 80 + x^{2} = 16$

Simplify.

Multiply $x$ by $x^{2}$ by adding the exponents.

Move $x^{2}$ .

$- (x^{2} x) - x (- 8 x) - x \cdot 16 + 5 x^{2} - 80 + x^{2} = 16$

Multiply $x^{2}$ by $x$ .

Raise $x$ to the power of $1$ .

$- (x^{2} x^{1}) - x (- 8 x) - x \cdot 16 + 5 x^{2} - 80 + x^{2} = 16$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- x^{2 + 1} - x (- 8 x) - x \cdot 16 + 5 x^{2} - 80 + x^{2} = 16$

Add $2$ and $1$ .

$- x^{3} - x (- 8 x) - x \cdot 16 + 5 x^{2} - 80 + x^{2} = 16$

Rewrite using the commutative property of multiplication.

$- x^{3} - 1 \cdot - 8 x \cdot x - x \cdot 16 + 5 x^{2} - 80 + x^{2} = 16$

Multiply $16$ by $- 1$ .

$- x^{3} - 1 \cdot - 8 x \cdot x - 16 x + 5 x^{2} - 80 + x^{2} = 16$

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$- x^{3} - 1 \cdot - 8 (x \cdot x) - 16 x + 5 x^{2} - 80 + x^{2} = 16$

Multiply $x$ by $x$ .

$- x^{3} - 1 \cdot - 8 x^{2} - 16 x + 5 x^{2} - 80 + x^{2} = 16$

Multiply $- 1$ by $- 8$ .

$- x^{3} + 8 x^{2} - 16 x + 5 x^{2} - 80 + x^{2} = 16$

Add $8 x^{2}$ and $5 x^{2}$ .

$- x^{3} + 13 x^{2} - 16 x - 80 + x^{2} = 16$

Add $13 x^{2}$ and $x^{2}$ .

$- x^{3} + 14 x^{2} - 16 x - 80 = 16$

Subtract $16$ from both sides of the equation.

$- x^{3} + 14 x^{2} - 16 x - 80 - 16 = 0$

Subtract $16$ from $- 80$ .

$- x^{3} + 14 x^{2} - 16 x - 96 = 0$

Factor the left side of the equation.

Factor $- x^{3} + 14 x^{2} - 16 x - 96$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 96, \pm 2, \pm 48, \pm 3, \pm 32, \pm 4, \pm 24, \pm 6, \pm 16, \pm 8, \pm 12 q = \pm 1$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 96, \pm 2, \pm 48, \pm 3, \pm 32, \pm 4, \pm 24, \pm 6, \pm 16, \pm 8, \pm 12$

Substitute $- 2$ and simplify the expression. In this case, the expression is equal to $0$ so $- 2$ is a root of the polynomial.

Substitute $- 2$ into the polynomial.

$- {(- 2)}^{3} + 14 {(- 2)}^{2} - 16 \cdot - 2 - 96$

Raise $- 2$ to the power of $3$ .

$- - 8 + 14 {(- 2)}^{2} - 16 \cdot - 2 - 96$

Multiply $- 1$ by $- 8$ .

$8 + 14 {(- 2)}^{2} - 16 \cdot - 2 - 96$

Raise $- 2$ to the power of $2$ .

$8 + 14 \cdot 4 - 16 \cdot - 2 - 96$

Multiply $14$ by $4$ .

$8 + 56 - 16 \cdot - 2 - 96$

Add $8$ and $56$ .

$64 - 16 \cdot - 2 - 96$

Multiply $- 16$ by $- 2$ .

$64 + 32 - 96$

Add $64$ and $32$ .

$96 - 96$

Subtract $96$ from $96$ .

$0$

Since $- 2$ is a known root, divide the polynomial by $x + 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- x^{3} + 14 x^{2} - 16 x - 96}{x + 2}$

Divide $- x^{3} + 14 x^{2} - 16 x - 96$ by $x + 2$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$2$

$x^{3}$

$14 x^{2}$

$16 x$

$96$

Divide the highest order term in the dividend $- x^{3}$ by the highest order term in divisor $x$ .

				-	$x^{2}$
	$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$

Multiply the new quotient term by the divisor.

			-	$x^{2}$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			-	$x^{3}$	-	$2 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $- x^{3} - 2 x^{2}$

			-	$x^{2}$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$
					+	$16 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

			-	$x^{2}$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$
					+	$16 x^{2}$	-	$16 x$

Divide the highest order term in the dividend $16 x^{2}$ by the highest order term in divisor $x$ .

			-	$x^{2}$	+	$16 x$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$
					+	$16 x^{2}$	-	$16 x$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$16 x$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$
					+	$16 x^{2}$	-	$16 x$
					+	$16 x^{2}$	+	$32 x$

The expression needs to be subtracted from the dividend, so change all the signs in $16 x^{2} + 32 x$

			-	$x^{2}$	+	$16 x$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$
					+	$16 x^{2}$	-	$16 x$
					-	$16 x^{2}$	-	$32 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$	+	$16 x$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$
					+	$16 x^{2}$	-	$16 x$
					-	$16 x^{2}$	-	$32 x$
							-	$48 x$

Pull the next terms from the original dividend down into the current dividend.

			-	$x^{2}$	+	$16 x$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$
					+	$16 x^{2}$	-	$16 x$
					-	$16 x^{2}$	-	$32 x$
							-	$48 x$	-	$96$

Divide the highest order term in the dividend $- 48 x$ by the highest order term in divisor $x$ .

			-	$x^{2}$	+	$16 x$	-	$48$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$
					+	$16 x^{2}$	-	$16 x$
					-	$16 x^{2}$	-	$32 x$
							-	$48 x$	-	$96$

Multiply the new quotient term by the divisor.

			-	$x^{2}$	+	$16 x$	-	$48$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$
					+	$16 x^{2}$	-	$16 x$
					-	$16 x^{2}$	-	$32 x$
							-	$48 x$	-	$96$
							-	$48 x$	-	$96$

The expression needs to be subtracted from the dividend, so change all the signs in $- 48 x - 96$

			-	$x^{2}$	+	$16 x$	-	$48$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$
					+	$16 x^{2}$	-	$16 x$
					-	$16 x^{2}$	-	$32 x$
							-	$48 x$	-	$96$
							+	$48 x$	+	$96$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$x^{2}$	+	$16 x$	-	$48$
$x$	+	$2$	-	$x^{3}$	+	$14 x^{2}$	-	$16 x$	-	$96$
			+	$x^{3}$	+	$2 x^{2}$
					+	$16 x^{2}$	-	$16 x$
					-	$16 x^{2}$	-	$32 x$
							-	$48 x$	-	$96$
							+	$48 x$	+	$96$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$- x^{2} + 16 x - 48$

Write $- x^{3} + 14 x^{2} - 16 x - 96$ as a set of factors.

$(x + 2) (- x^{2} + 16 x - 48) = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 48 = 48$ and whose sum is $b = 16$ .

Factor $16$ out of $16 x$ .

$(x + 2) (- x^{2} + 16 (x) - 48) = 0$

Rewrite $16$ as $4$ plus $12$

$(x + 2) (- x^{2} + (4 + 12) x - 48) = 0$

Apply the distributive property.

$(x + 2) (- x^{2} + 4 x + 12 x - 48) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 2) ((- x^{2} + 4 x) + 12 x - 48) = 0$

Factor out the greatest common factor (GCF) from each group.

$(x + 2) (x (- x + 4) - 12 (- x + 4)) = 0$

Factor the polynomial by factoring out the greatest common factor, $- x + 4$ .

$(x + 2) ((- x + 4) (x - 12)) = 0$

Remove unnecessary parentheses.

$(x + 2) (- x + 4) (x - 12) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 2 = 0$

$- x + 4 = 0$

$x - 12 = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $- x + 4$ equal to $0$ and solve for $x$ .

Set $- x + 4$ equal to $0$ .

$- x + 4 = 0$

Solve $- x + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$- x = - 4$

Divide each term in $- x = - 4$ by $- 1$ and simplify.

Divide each term in $- x = - 4$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x = 4$

Set $x - 12$ equal to $0$ and solve for $x$ .

Set $x - 12$ equal to $0$ .

$x - 12 = 0$

Add $12$ to both sides of the equation.

$x = 12$

The final solution is all the values that make $(x + 2) (- x + 4) (x - 12) = 0$ true.

$x = - 2, 4, 12$

Step 4

Exclude the solutions that do not make $\frac{x}{x + 4} + \frac{5}{4 - x} = \frac{1}{x - 4}$ true.

$x = - 2, 12$

Solve for x x/(x+4)+5/(4-x)=1/(x-4)

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