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

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

Multiply the numerator of the first fraction by the denominator of the second fraction. Set this equal to the product of the denominator of the first fraction and the numerator of the second fraction.

$x (x + 4) = 5 \cdot 1$

Solve the equation for $x$ .

Simplify $x (x + 4)$ .

Apply the distributive property.

$x \cdot x + x \cdot 4 = 5 \cdot 1$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 4 = 5 \cdot 1$

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

$x^{2} + 4 x = 5 \cdot 1$

Multiply $5$ by $1$ .

$x^{2} + 4 x = 5$

Subtract $5$ from both sides of the equation.

$x^{2} + 4 x - 5 = 0$

Factor $x^{2} + 4 x - 5$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

$(x - 1) (x + 5) = 0$

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

$x - 1 = 0$

$x + 5 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 1) (x + 5) = 0$ true.

$x = 1, - 5$

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

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