Algebra Examples

$- \frac{4}{j + 1} = j - 3$

Step 1

Find the LCD of the terms in the equation.

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Step 1.1

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

$j + 1, 1, 1$

Step 1.2

Remove parentheses.

$j + 1, 1, 1$

Step 1.3

The LCM of one and any expression is the expression.

$j + 1$

Step 2

Multiply each term in $- \frac{4}{j + 1} = j - 3$ by $j + 1$ to eliminate the fractions.

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Step 2.1

Multiply each term in $- \frac{4}{j + 1} = j - 3$ by $j + 1$ .

$- \frac{4}{j + 1} (j + 1) = j (j + 1) - 3 (j + 1)$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $j + 1$ .

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Step 2.2.1.1

Move the leading negative in $- \frac{4}{j + 1}$ into the numerator.

$\frac{- 4}{j + 1} (j + 1) = j (j + 1) - 3 (j + 1)$

Step 2.2.1.2

Cancel the common factor.

$\frac{- 4}{j + 1} (j + 1) = j (j + 1) - 3 (j + 1)$

Step 2.2.1.3

Rewrite the expression.

$- 4 = j (j + 1) - 3 (j + 1)$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Apply the distributive property.

$- 4 = j \cdot j + j \cdot 1 - 3 (j + 1)$

Step 2.3.1.2

Multiply $j$ by $j$ .

$- 4 = j^{2} + j \cdot 1 - 3 (j + 1)$

Step 2.3.1.3

Multiply $j$ by $1$ .

$- 4 = j^{2} + j - 3 (j + 1)$

Step 2.3.1.4

Apply the distributive property.

$- 4 = j^{2} + j - 3 j - 3 \cdot 1$

Step 2.3.1.5

Multiply $- 3$ by $1$ .

$- 4 = j^{2} + j - 3 j - 3$

Step 2.3.2

Subtract $3 j$ from $j$ .

$- 4 = j^{2} - 2 j - 3$

Step 3

Solve the equation.

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Step 3.1

Rewrite the equation as $j^{2} - 2 j - 3 = - 4$ .

$j^{2} - 2 j - 3 = - 4$

Step 3.2

Add $4$ to both sides of the equation.

$j^{2} - 2 j - 3 + 4 = 0$

Step 3.3

Add $- 3$ and $4$ .

$j^{2} - 2 j + 1 = 0$

Step 3.4

Factor using the perfect square rule.

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Step 3.4.1

Rewrite $1$ as $1^{2}$ .

$j^{2} - 2 j + 1^{2} = 0$

Step 3.4.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 j = 2 \cdot j \cdot 1$

Step 3.4.3

Rewrite the polynomial.

$j^{2} - 2 \cdot j \cdot 1 + 1^{2} = 0$

Step 3.4.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = j$ and $b = 1$ .

${(j - 1)}^{2} = 0$

Step 3.5

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

$j - 1 = 0$

Step 3.6

Add $1$ to both sides of the equation.

$j = 1$