Algebra Examples

\frac{k + 4}{4} + \frac{k - 1}{4} = \frac{k + 4}{4 k}

$\frac{k + 4}{4} + \frac{k - 1}{4} = \frac{k + 4}{4 k}$

Step 1

Subtract

\frac{k + 4}{4 k}

$\frac{k + 4}{4 k}$ from both sides of the equation.

\frac{k + 4}{4} + \frac{k - 1}{4} - \frac{k + 4}{4 k} = 0

$\frac{k + 4}{4} + \frac{k - 1}{4} - \frac{k + 4}{4 k} = 0$

Step 2

Simplify

\frac{k + 4}{4} + \frac{k - 1}{4} - \frac{k + 4}{4 k}

$\frac{k + 4}{4} + \frac{k - 1}{4} - \frac{k + 4}{4 k}$ .

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Step 2.1

Combine the numerators over the common denominator.

\frac{k + 4 + k - 1}{4} + \frac{- (k + 4)}{4 k} = 0

$\frac{k + 4 + k - 1}{4} + \frac{- (k + 4)}{4 k} = 0$

Step 2.2

Add

k

$k$ and

k

$k$ .

\frac{2 k + 4 - 1}{4} + \frac{- (k + 4)}{4 k} = 0

$\frac{2 k + 4 - 1}{4} + \frac{- (k + 4)}{4 k} = 0$

Step 2.3

Subtract

1

$1$ from

4

$4$ .

\frac{2 k + 3}{4} + \frac{- (k + 4)}{4 k} = 0

$\frac{2 k + 3}{4} + \frac{- (k + 4)}{4 k} = 0$

Step 2.4

Move the negative in front of the fraction.

$\frac{2 k + 3}{4} - \frac{k + 4}{4 k} = 0$

Step 2.5

To write

$\frac{2 k + 3}{4}$ as a fraction with a common denominator, multiply by

$\frac{k}{k}$ .

$\frac{2 k + 3}{4} \cdot \frac{k}{k} - \frac{k + 4}{4 k} = 0$

Step 2.6

Multiply

$\frac{2 k + 3}{4}$ by

$\frac{k}{k}$ .

$\frac{(2 k + 3) k}{4 k} - \frac{k + 4}{4 k} = 0$

Step 2.7

Combine the numerators over the common denominator.

$\frac{(2 k + 3) k - (k + 4)}{4 k} = 0$

Step 2.8

Simplify the numerator.

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Step 2.8.1

Apply the distributive property.

$\frac{2 k \cdot k + 3 k - (k + 4)}{4 k} = 0$

Step 2.8.2

Multiply

$k$ by

$k$ by adding the exponents.

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Step 2.8.2.1

Move

$k$ .

$\frac{2 (k \cdot k) + 3 k - (k + 4)}{4 k} = 0$

Step 2.8.2.2

Multiply

$k$ by

$k$ .

$\frac{2 k^{2} + 3 k - (k + 4)}{4 k} = 0$

Step 2.8.3

Apply the distributive property.

$\frac{2 k^{2} + 3 k - k - 1 \cdot 4}{4 k} = 0$

Step 2.8.4

Multiply

$- 1$ by

$4$ .

$\frac{2 k^{2} + 3 k - k - 4}{4 k} = 0$

Step 2.8.5

Subtract

$k$ from

$3 k$ .

$\frac{2 k^{2} + 2 k - 4}{4 k} = 0$

Step 2.8.6

Rewrite

$2 k^{2} + 2 k - 4$ in a factored form.

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Step 2.8.6.1

Factor

$2$ out of

$2 k^{2} + 2 k - 4$ .

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Step 2.8.6.1.1

Factor

$2$ out of

$2 k^{2}$ .

$\frac{2 (k^{2}) + 2 k - 4}{4 k} = 0$

Step 2.8.6.1.2

Factor

$2$ out of

$2 k$ .

$\frac{2 (k^{2}) + 2 (k) - 4}{4 k} = 0$

Step 2.8.6.1.3

Factor

$2$ out of

$- 4$ .

$\frac{2 (k^{2}) + 2 k + 2 \cdot - 2}{4 k} = 0$

Step 2.8.6.1.4

Factor

$2$ out of

$2 (k^{2}) + 2 k$ .

$\frac{2 (k^{2} + k) + 2 \cdot - 2}{4 k} = 0$

Step 2.8.6.1.5

Factor

$2$ out of

$2 (k^{2} + k) + 2 \cdot - 2$ .

$\frac{2 (k^{2} + k - 2)}{4 k} = 0$

Step 2.8.6.2

Factor

$k^{2} + k - 2$ using the AC method.

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Step 2.8.6.2.1

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

$- 2$ and whose sum is

$1$ .

$- 1, 2$

Step 2.8.6.2.2

Write the factored form using these integers.

$\frac{2 ((k - 1) (k + 2))}{4 k} = 0$

$\frac{2 (k - 1) (k + 2)}{4 k} = 0$

Step 2.9

Cancel the common factor of

$2$ and

$4$ .

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Step 2.9.1

Factor

$2$ out of

$2 (k - 1) (k + 2)$ .

$\frac{2 ((k - 1) (k + 2))}{4 k} = 0$

Step 2.9.2

Cancel the common factors.

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Step 2.9.2.1

Factor

$2$ out of

$4 k$ .

$\frac{2 ((k - 1) (k + 2))}{2 (2 k)} = 0$

Step 2.9.2.2

Cancel the common factor.

$\frac{2 ((k - 1) (k + 2))}{2 (2 k)} = 0$

Step 2.9.2.3

Rewrite the expression.

$\frac{(k - 1) (k + 2)}{2 k} = 0$

Step 3

Set the numerator equal to zero.

$(k - 1) (k + 2) = 0$

Step 4

Solve the equation for

$k$ .

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Step 4.1

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$k - 1 = 0$

$k + 2 = 0$

Step 4.2

Set

$k - 1$ equal to

$0$ and solve for

$k$ .

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Step 4.2.1

Set

$k - 1$ equal to

$0$ .

$k - 1 = 0$

Step 4.2.2

Add

$1$ to both sides of the equation.

$k = 1$

Step 4.3

Set

$k + 2$ equal to

$0$ and solve for

$k$ .

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Step 4.3.1

Set

$k + 2$ equal to

$0$ .

$k + 2 = 0$

Step 4.3.2

Subtract

$2$ from both sides of the equation.

$k = - 2$

Step 4.4

The final solution is all the values that make

$(k - 1) (k + 2) = 0$ true.

$k = 1, - 2$