Solve for k (k+4)/4+(k-1)/4=(k+4)/(4k)

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

Find the LCD of the terms in the equation.

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

$4, 4, 4 k$

Since $4, 4, 4 k$ contain both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $4, 4, 4$ then find LCM for the variable part $k^{1}$ .

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

$4$ has factors of $2$ and $2$ .

$2 \cdot 2$

Multiply $2$ by $2$ .

$4$

The factor for $k^{1}$ is $k$ itself.

$k^{1} = k$

$k$ occurs $1$ time.

The LCM of $k^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$k$

The LCM for $4, 4, 4 k$ is the numeric part $4$ multiplied by the variable part.

$4 k$

Multiply each term by $4 k$ and simplify.

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Multiply each term in $\frac{k + 4}{4} + \frac{k - 1}{4} = \frac{k + 4}{4 k}$ by $4 k$ in order to remove all the denominators from the equation.

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

Simplify $\frac{k + 4}{4} \cdot (4 k) + \frac{k - 1}{4} \cdot (4 k)$ .

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Simplify each term.

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Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $4$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply $k$ by $k$ .

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

Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $4$ .

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Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply $k$ by $k$ .

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

Rewrite $- 1 k$ as $- k$ .

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

Simplify by adding terms.

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Add $k^{2}$ and $k^{2}$ .

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

Subtract $k$ from $4 k$ .

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

Simplify $\frac{k + 4}{4 k} \cdot (4 k)$ .

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Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $4$ .

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Factor $4$ out of $4 k$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $k$ .

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Cancel the common factor.

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

Rewrite the expression.

$2 k^{2} + 3 k = k + 4$

Solve the equation.

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Move all terms containing $k$ to the left side of the equation.

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Subtract $k$ from both sides of the equation.

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

Subtract $k$ from $3 k$ .

$2 k^{2} + 2 k = 4$

Move $4$ to the left side of the equation by subtracting it from both sides.

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

Factor the left side of the equation.

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Factor $2$ out of $2 k^{2} + 2 k - 4$ .

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Factor $2$ out of $2 k^{2}$ .

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

Factor $2$ out of $2 k$ .

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

Factor $2$ out of $- 4$ .

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

Factor $2$ out of $2 (k^{2}) + 2 k$ .

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

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

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

Factor.

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

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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 $1$ .

$- 1, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

Divide each term by $2$ and simplify.

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Divide each term in $2 (k - 1) (k + 2) = 0$ by $2$ .

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

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

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

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

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

Expand $(k - 1) (k + 2)$ using the FOIL Method.

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Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

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Simplify each term.

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Multiply $k$ by $k$ .

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

Move $2$ to the left of $k$ .

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

Rewrite $- 1 k$ as $- k$ .

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

Multiply $- 1$ by $2$ .

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

Subtract $k$ from $2 k$ .

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

Divide $0$ by $2$ .

$k^{2} + k - 2 = 0$

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

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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 $1$ .

$- 1, 2$

Write the factored form using these integers.

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

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

$k - 1 = 0$

$k + 2 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$k - 1 = 0$

Add $1$ to both sides of the equation.

$k = 1$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$k + 2 = 0$

Subtract $2$ from both sides of the equation.

$k = - 2$

The final solution is all the values that make $(k - 1) (k + 2) = 0$ true.

$k = 1, - 2$

Solve for k (k+4)/4+(k-1)/4=(k+4)/(4k)

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