Solve for k 1/(2k)+4/k=k/(k+2)

$\frac{1}{2 k} + \frac{4}{k} = \frac{k}{k + 2}$

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.

$2 k, k, k + 2$

Since $2 k, k, k + 2$ contain both numbers and variables, there are four steps to find the LCM. Find LCM for the numeric, variable, and compound variable parts. Then, multiply them all together.

Steps to find the LCM for $2 k, k, k + 2$ are:

1. Find the LCM for the numeric part $2, 1, 1$ .

2. Find the LCM for the variable part $k^{1}, k^{1}$ .

3. Find the LCM for the compound variable part $k + 2$ .

4. Multiply each LCM together.

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.

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of $2, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2$

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

$k^{1} = k$

$k$ occurs $1$ time.

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

$k$

The factor for $k + 2$ is $k + 2$ itself.

$(k + 2) = k + 2$

$(k + 2)$ occurs $1$ time.

The LCM of $k + 2$ is the result of multiplying all factors the greatest number of times they occur in either term.

$k + 2$

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$2 k (k + 2)$

Multiply each term in $\frac{1}{2 k} + \frac{4}{k} = \frac{k}{k + 2}$ by $2 k (k + 2)$ to eliminate the fractions.

Multiply each term in $\frac{1}{2 k} + \frac{4}{k} = \frac{k}{k + 2}$ by $2 k (k + 2)$ .

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

Simplify the left side.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 k$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $k$ .

Cancel the common factor.

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

Rewrite the expression.

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

Rewrite using the commutative property of multiplication.

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

Multiply $2 \frac{4}{k}$ .

Combine $2$ and $\frac{4}{k}$ .

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

Multiply $2$ by $4$ .

$k + 2 + \frac{8}{k} (k (k + 2)) = \frac{k}{k + 2} (2 k (k + 2))$

Cancel the common factor of $k$ .

Cancel the common factor.

$k + 2 + \frac{8}{k} (k (k + 2)) = \frac{k}{k + 2} (2 k (k + 2))$

Rewrite the expression.

$k + 2 + 8 (k + 2) = \frac{k}{k + 2} (2 k (k + 2))$

Apply the distributive property.

$k + 2 + 8 k + 8 \cdot 2 = \frac{k}{k + 2} (2 k (k + 2))$

Multiply $8$ by $2$ .

$k + 2 + 8 k + 16 = \frac{k}{k + 2} (2 k (k + 2))$

Simplify by adding terms.

Add $k$ and $8 k$ .

$9 k + 2 + 16 = \frac{k}{k + 2} (2 k (k + 2))$

Add $2$ and $16$ .

$9 k + 18 = \frac{k}{k + 2} (2 k (k + 2))$

Simplify the right side.

Rewrite using the commutative property of multiplication.

$9 k + 18 = 2 \frac{k}{k + 2} (k (k + 2))$

Combine $2$ and $\frac{k}{k + 2}$ .

$9 k + 18 = \frac{2 k}{k + 2} (k (k + 2))$

Cancel the common factor of $k + 2$ .

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

$9 k + 18 = \frac{2 k}{k + 2} ((k + 2) k)$

Cancel the common factor.

$9 k + 18 = \frac{2 k}{k + 2} ((k + 2) k)$

Rewrite the expression.

$9 k + 18 = 2 k \cdot k$

Raise $k$ to the power of $1$ .

$9 k + 18 = 2 (k^{1} k)$

Raise $k$ to the power of $1$ .

$9 k + 18 = 2 (k^{1} k^{1})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$9 k + 18 = 2 k^{1 + 1}$

Add $1$ and $1$ .

$9 k + 18 = 2 k^{2}$

Solve the equation.

Subtract $2 k^{2}$ from both sides of the equation.

$9 k + 18 - 2 k^{2} = 0$

Factor the left side of the equation.

Factor $- 1$ out of $9 k + 18 - 2 k^{2}$ .

Reorder the expression.

Move $18$ .

$9 k - 2 k^{2} + 18 = 0$

Reorder $9 k$ and $- 2 k^{2}$ .

$- 2 k^{2} + 9 k + 18 = 0$

Factor $- 1$ out of $- 2 k^{2}$ .

$- (2 k^{2}) + 9 k + 18 = 0$

Factor $- 1$ out of $9 k$ .

$- (2 k^{2}) - (- 9 k) + 18 = 0$

Rewrite $18$ as $- 1 (- 18)$ .

$- (2 k^{2}) - (- 9 k) - 1 \cdot - 18 = 0$

Factor $- 1$ out of $- (2 k^{2}) - (- 9 k)$ .

$- (2 k^{2} - 9 k) - 1 \cdot - 18 = 0$

Factor $- 1$ out of $- (2 k^{2} - 9 k) - 1 (- 18)$ .

$- (2 k^{2} - 9 k - 18) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 18 = - 36$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 k$ .

$- (2 k^{2} - 9 k - 18) = 0$

Rewrite $- 9$ as $3$ plus $- 12$

$- (2 k^{2} + (3 - 12) k - 18) = 0$

Apply the distributive property.

$- (2 k^{2} + 3 k - 12 k - 18) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((2 k^{2} + 3 k) - 12 k - 18) = 0$

Factor out the greatest common factor (GCF) from each group.

$- (k (2 k + 3) - 6 (2 k + 3)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 k + 3$ .

$- ((2 k + 3) (k - 6)) = 0$

Remove unnecessary parentheses.

$- (2 k + 3) (k - 6) = 0$

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

$2 k + 3 = 0$

$k - 6 = 0$

Set $2 k + 3$ equal to $0$ and solve for $k$ .

Set $2 k + 3$ equal to $0$ .

$2 k + 3 = 0$

Solve $2 k + 3 = 0$ for $k$ .

Subtract $3$ from both sides of the equation.

$2 k = - 3$

Divide each term in $2 k = - 3$ by $2$ and simplify.

Divide each term in $2 k = - 3$ by $2$ .

$\frac{2 k}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 k}{2} = \frac{- 3}{2}$

Divide $k$ by $1$ .

$k = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$k = - \frac{3}{2}$

Set $k - 6$ equal to $0$ and solve for $k$ .

Set $k - 6$ equal to $0$ .

$k - 6 = 0$

Add $6$ to both sides of the equation.

$k = 6$

The final solution is all the values that make $- (2 k + 3) (k - 6) = 0$ true.

$k = - \frac{3}{2}, 6$

The result can be shown in multiple forms.

Exact Form:

$k = - \frac{3}{2}, 6$

Decimal Form:

$k = - 1.5, 6$

Mixed Number Form:

$k = - 1 \frac{1}{2}, 6$

Solve for k 1/(2k)+4/k=k/(k+2)

Related Questions

Support LCMGCF.com by disabling your adblocker.