Solve for j 2/(j-5)=j-6

$\frac{2}{j - 5} = j - 6$

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.

$j - 5, 1, 1$

Remove parentheses.

$j - 5, 1, 1$

The LCM of one and any expression is the expression.

$j - 5$

Step 2

Multiply each term in $\frac{2}{j - 5} = j - 6$ by $j - 5$ to eliminate the fractions.

Multiply each term in $\frac{2}{j - 5} = j - 6$ by $j - 5$ .

$\frac{2}{j - 5} (j - 5) = j (j - 5) - 6 (j - 5)$

Simplify the left side.

Cancel the common factor of $j - 5$ .

Cancel the common factor.

$\frac{2}{j - 5} (j - 5) = j (j - 5) - 6 (j - 5)$

Rewrite the expression.

$2 = j (j - 5) - 6 (j - 5)$

Simplify the right side.

Simplify each term.

Apply the distributive property.

$2 = j \cdot j + j \cdot - 5 - 6 (j - 5)$

Multiply $j$ by $j$ .

$2 = j^{2} + j \cdot - 5 - 6 (j - 5)$

Move $- 5$ to the left of $j$ .

$2 = j^{2} - 5 \cdot j - 6 (j - 5)$

Apply the distributive property.

$2 = j^{2} - 5 j - 6 j - 6 \cdot - 5$

Multiply $- 6$ by $- 5$ .

$2 = j^{2} - 5 j - 6 j + 30$

Subtract $6 j$ from $- 5 j$ .

$2 = j^{2} - 11 j + 30$

Step 3

Solve the equation.

Rewrite the equation as $j^{2} - 11 j + 30 = 2$ .

$j^{2} - 11 j + 30 = 2$

Subtract $2$ from both sides of the equation.

$j^{2} - 11 j + 30 - 2 = 0$

Subtract $2$ from $30$ .

$j^{2} - 11 j + 28 = 0$

Factor $j^{2} - 11 j + 28$ 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 $28$ and whose sum is $- 11$ .

$- 7, - 4$

Write the factored form using these integers.

$(j - 7) (j - 4) = 0$

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

$j - 7 = 0$

$j - 4 = 0$

Set $j - 7$ equal to $0$ and solve for $j$ .

Set $j - 7$ equal to $0$ .

$j - 7 = 0$

Add $7$ to both sides of the equation.

$j = 7$

Set $j - 4$ equal to $0$ and solve for $j$ .

Set $j - 4$ equal to $0$ .

$j - 4 = 0$

Add $4$ to both sides of the equation.

$j = 4$

The final solution is all the values that make $(j - 7) (j - 4) = 0$ true.

$j = 7, 4$

Solve for j 2/(j-5)=j-6

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