Pre-Algebra Examples

$\frac{- 4}{2 k + 2} + \frac{7}{2 k - 2} = \frac{7}{4 k^{2 - 4}}$

Step 1

Simplify $\frac{- 4}{2 k + 2} + \frac{7}{2 k - 2}$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Cancel the common factor of $- 4$ and $2 k + 2$ .

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Step 1.1.1.1

Factor $2$ out of $- 4$ .

$\frac{2 \cdot - 2}{2 k + 2} + \frac{7}{2 k - 2} = \frac{7}{4 k^{2 - 4}}$

Step 1.1.1.2

Cancel the common factors.

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Step 1.1.1.2.1

Factor $2$ out of $2 k$ .

$\frac{2 \cdot - 2}{2 (k) + 2} + \frac{7}{2 k - 2} = \frac{7}{4 k^{2 - 4}}$

Step 1.1.1.2.2

Factor $2$ out of $2$ .

$\frac{2 \cdot - 2}{2 (k) + 2 (1)} + \frac{7}{2 k - 2} = \frac{7}{4 k^{2 - 4}}$

Step 1.1.1.2.3

Factor $2$ out of $2 (k) + 2 (1)$ .

$\frac{2 \cdot - 2}{2 (k + 1)} + \frac{7}{2 k - 2} = \frac{7}{4 k^{2 - 4}}$

Step 1.1.1.2.4

Cancel the common factor.

$\frac{2 \cdot - 2}{2 (k + 1)} + \frac{7}{2 k - 2} = \frac{7}{4 k^{2 - 4}}$

Step 1.1.1.2.5

Rewrite the expression.

$\frac{- 2}{k + 1} + \frac{7}{2 k - 2} = \frac{7}{4 k^{2 - 4}}$

Step 1.1.2

Move the negative in front of the fraction.

$- \frac{2}{k + 1} + \frac{7}{2 k - 2} = \frac{7}{4 k^{2 - 4}}$

Step 1.1.3

Factor $2$ out of $2 k - 2$ .

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Step 1.1.3.1

Factor $2$ out of $2 k$ .

$- \frac{2}{k + 1} + \frac{7}{2 (k) - 2} = \frac{7}{4 k^{2 - 4}}$

Step 1.1.3.2

Factor $2$ out of $- 2$ .

$- \frac{2}{k + 1} + \frac{7}{2 (k) + 2 (- 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.1.3.3

Factor $2$ out of $2 (k) + 2 (- 1)$ .

$- \frac{2}{k + 1} + \frac{7}{2 (k - 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.2

To write $- \frac{2}{k + 1}$ as a fraction with a common denominator, multiply by $\frac{2 (k - 1)}{2 (k - 1)}$ .

$- \frac{2}{k + 1} \cdot \frac{2 (k - 1)}{2 (k - 1)} + \frac{7}{2 (k - 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.3

To write $\frac{7}{2 (k - 1)}$ as a fraction with a common denominator, multiply by $\frac{k + 1}{k + 1}$ .

$- \frac{2}{k + 1} \cdot \frac{2 (k - 1)}{2 (k - 1)} + \frac{7}{2 (k - 1)} \cdot \frac{k + 1}{k + 1} = \frac{7}{4 k^{2 - 4}}$

Step 1.4

Write each expression with a common denominator of $(k + 1) \cdot 2 (k - 1)$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.4.1

Multiply $\frac{2}{k + 1}$ by $\frac{2 (k - 1)}{2 (k - 1)}$ .

$- \frac{2 (2 (k - 1))}{(k + 1) (2 (k - 1))} + \frac{7}{2 (k - 1)} \cdot \frac{k + 1}{k + 1} = \frac{7}{4 k^{2 - 4}}$

Step 1.4.2

Multiply $\frac{7}{2 (k - 1)}$ by $\frac{k + 1}{k + 1}$ .

$- \frac{2 (2 (k - 1))}{(k + 1) (2 (k - 1))} + \frac{7 (k + 1)}{2 (k - 1) (k + 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.4.3

Reorder the factors of $(k + 1) (2 (k - 1))$ .

$- \frac{2 (2 (k - 1))}{2 (k + 1) (k - 1)} + \frac{7 (k + 1)}{2 (k - 1) (k + 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.4.4

Reorder the factors of $2 (k - 1) (k + 1)$ .

$- \frac{2 (2 (k - 1))}{2 (k + 1) (k - 1)} + \frac{7 (k + 1)}{2 (k + 1) (k - 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.5

Combine the numerators over the common denominator.

$\frac{- 2 (2 (k - 1)) + 7 (k + 1)}{2 (k + 1) (k - 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.6

Simplify the numerator.

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Step 1.6.1

Multiply $- 2$ by $2$ .

$\frac{- 4 (k - 1) + 7 (k + 1)}{2 (k + 1) (k - 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.6.2

Apply the distributive property.

$\frac{- 4 k - 4 \cdot - 1 + 7 (k + 1)}{2 (k + 1) (k - 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.6.3

Multiply $- 4$ by $- 1$ .

$\frac{- 4 k + 4 + 7 (k + 1)}{2 (k + 1) (k - 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.6.4

Apply the distributive property.

$\frac{- 4 k + 4 + 7 k + 7 \cdot 1}{2 (k + 1) (k - 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.6.5

Multiply $7$ by $1$ .

$\frac{- 4 k + 4 + 7 k + 7}{2 (k + 1) (k - 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.6.6

Add $- 4 k$ and $7 k$ .

$\frac{3 k + 4 + 7}{2 (k + 1) (k - 1)} = \frac{7}{4 k^{2 - 4}}$

Step 1.6.7

Add $4$ and $7$ .

$\frac{3 k + 11}{2 (k + 1) (k - 1)} = \frac{7}{4 k^{2 - 4}}$

Step 2

Simplify $\frac{7}{4 k^{2 - 4}}$ .

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Step 2.1

Move $k^{2 - 4}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$\frac{3 k + 11}{2 (k + 1) (k - 1)} = \frac{7 k^{- (2 - 4)}}{4}$

Step 2.2

Simplify the numerator.

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Step 2.2.1

Subtract $4$ from $2$ .

$\frac{3 k + 11}{2 (k + 1) (k - 1)} = \frac{7 k^{- - 2}}{4}$

Step 2.2.2

Multiply $- 1$ by $- 2$ .

$\frac{3 k + 11}{2 (k + 1) (k - 1)} = \frac{7 k^{2}}{4}$

Step 3

Graph each side of the equation. The solution is the x-value of the point of intersection.

$k \approx - 1.40672727, 1.64031257$

Step 4