Solve for j -5/(j+4)=j-2

$- \frac{5}{j + 4} = j - 2$

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.

$j + 4, 1, 1$

Remove parentheses.

$j + 4, 1, 1$

The LCM of one and any expression is the expression.

$j + 4$

Multiply each term in $- \frac{5}{j + 4} = j - 2$ by $j + 4$ to eliminate the fractions.

Multiply each term in $- \frac{5}{j + 4} = j - 2$ by $j + 4$ .

$- \frac{5}{j + 4} (j + 4) = j (j + 4) - 2 (j + 4)$

Simplify the left side.

Cancel the common factor of $j + 4$ .

Move the leading negative in $- \frac{5}{j + 4}$ into the numerator.

$\frac{- 5}{j + 4} (j + 4) = j (j + 4) - 2 (j + 4)$

Cancel the common factor.

$\frac{- 5}{j + 4} (j + 4) = j (j + 4) - 2 (j + 4)$

Rewrite the expression.

$- 5 = j (j + 4) - 2 (j + 4)$

Simplify the right side.

Simplify each term.

Apply the distributive property.

$- 5 = j \cdot j + j \cdot 4 - 2 (j + 4)$

Multiply $j$ by $j$ .

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

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

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

Apply the distributive property.

$- 5 = j^{2} + 4 j - 2 j - 2 \cdot 4$

Multiply $- 2$ by $4$ .

$- 5 = j^{2} + 4 j - 2 j - 8$

Subtract $2 j$ from $4 j$ .

$- 5 = j^{2} + 2 j - 8$

Solve the equation.

Rewrite the equation as $j^{2} + 2 j - 8 = - 5$ .

$j^{2} + 2 j - 8 = - 5$

Add $5$ to both sides of the equation.

$j^{2} + 2 j - 8 + 5 = 0$

Add $- 8$ and $5$ .

$j^{2} + 2 j - 3 = 0$

Factor $j^{2} + 2 j - 3$ 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 $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$(j - 1) (j + 3) = 0$

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

$j - 1 = 0$

$j + 3 = 0$

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

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

$j - 1 = 0$

Add $1$ to both sides of the equation.

$j = 1$

Set $j + 3$ equal to $0$ and solve for $j$ .

Set $j + 3$ equal to $0$ .

$j + 3 = 0$

Subtract $3$ from both sides of the equation.

$j = - 3$

The final solution is all the values that make $(j - 1) (j + 3) = 0$ true.

$j = 1, - 3$

Solve for j -5/(j+4)=j-2

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