Solve for k 5/4(2-k)=2(3k-1)-2/3k

$\frac{5}{4} (2 - k) = 2 (3 k - 1) - \frac{2}{3} k$

Step 1

Since $k$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$2 (3 k - 1) - \frac{2}{3} k = \frac{5}{4} \cdot (2 - k)$

Step 2

Simplify $2 (3 k - 1) - \frac{2}{3} k$ .

Simplify each term.

Apply the distributive property.

$2 (3 k) + 2 \cdot - 1 - \frac{2}{3} k = \frac{5}{4} \cdot (2 - k)$

Multiply $3$ by $2$ .

$6 k + 2 \cdot - 1 - \frac{2}{3} k = \frac{5}{4} \cdot (2 - k)$

Multiply $2$ by $- 1$ .

$6 k - 2 - \frac{2}{3} k = \frac{5}{4} \cdot (2 - k)$

Combine $k$ and $\frac{2}{3}$ .

$6 k - 2 - \frac{k \cdot 2}{3} = \frac{5}{4} \cdot (2 - k)$

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

$6 k - 2 - \frac{2 k}{3} = \frac{5}{4} \cdot (2 - k)$

To write $6 k$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$6 k \cdot \frac{3}{3} - \frac{2 k}{3} - 2 = \frac{5}{4} \cdot (2 - k)$

Simplify terms.

Combine $6 k$ and $\frac{3}{3}$ .

$\frac{6 k \cdot 3}{3} - \frac{2 k}{3} - 2 = \frac{5}{4} \cdot (2 - k)$

Combine the numerators over the common denominator.

$\frac{6 k \cdot 3 - 2 k}{3} - 2 = \frac{5}{4} \cdot (2 - k)$

Simplify the numerator.

Factor $2 k$ out of $6 k \cdot 3 - 2 k$ .

Factor $2 k$ out of $6 k \cdot 3$ .

$\frac{2 k (3 \cdot 3) - 2 k}{3} - 2 = \frac{5}{4} \cdot (2 - k)$

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

$\frac{2 k (3 \cdot 3) + 2 k (- 1)}{3} - 2 = \frac{5}{4} \cdot (2 - k)$

Factor $2 k$ out of $2 k (3 \cdot 3) + 2 k (- 1)$ .

$\frac{2 k (3 \cdot 3 - 1)}{3} - 2 = \frac{5}{4} \cdot (2 - k)$

Multiply $3$ by $3$ .

$\frac{2 k (9 - 1)}{3} - 2 = \frac{5}{4} \cdot (2 - k)$

Subtract $1$ from $9$ .

$\frac{2 k \cdot 8}{3} - 2 = \frac{5}{4} \cdot (2 - k)$

Multiply $8$ by $2$ .

$\frac{16 k}{3} - 2 = \frac{5}{4} \cdot (2 - k)$

Step 3

Simplify $\frac{5}{4} \cdot (2 - k)$ .

Apply the distributive property.

$\frac{16 k}{3} - 2 = \frac{5}{4} \cdot 2 + \frac{5}{4} (- k)$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$\frac{16 k}{3} - 2 = \frac{5}{2 (2)} \cdot 2 + \frac{5}{4} (- k)$

Cancel the common factor.

$\frac{16 k}{3} - 2 = \frac{5}{2 \cdot 2} \cdot 2 + \frac{5}{4} (- k)$

Rewrite the expression.

$\frac{16 k}{3} - 2 = \frac{5}{2} + \frac{5}{4} (- k)$

Combine $\frac{5}{4}$ and $k$ .

$\frac{16 k}{3} - 2 = \frac{5}{2} - \frac{5 k}{4}$

Step 4

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

Add $\frac{5 k}{4}$ to both sides of the equation.

$\frac{16 k}{3} - 2 + \frac{5 k}{4} = \frac{5}{2}$

To write $\frac{16 k}{3}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{16 k}{3} \cdot \frac{4}{4} + \frac{5 k}{4} - 2 = \frac{5}{2}$

To write $\frac{5 k}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{16 k}{3} \cdot \frac{4}{4} + \frac{5 k}{4} \cdot \frac{3}{3} - 2 = \frac{5}{2}$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{16 k}{3}$ by $\frac{4}{4}$ .

$\frac{16 k \cdot 4}{3 \cdot 4} + \frac{5 k}{4} \cdot \frac{3}{3} - 2 = \frac{5}{2}$

Multiply $3$ by $4$ .

$\frac{16 k \cdot 4}{12} + \frac{5 k}{4} \cdot \frac{3}{3} - 2 = \frac{5}{2}$

Multiply $\frac{5 k}{4}$ by $\frac{3}{3}$ .

$\frac{16 k \cdot 4}{12} + \frac{5 k \cdot 3}{4 \cdot 3} - 2 = \frac{5}{2}$

Multiply $4$ by $3$ .

$\frac{16 k \cdot 4}{12} + \frac{5 k \cdot 3}{12} - 2 = \frac{5}{2}$

Combine the numerators over the common denominator.

$\frac{16 k \cdot 4 + 5 k \cdot 3}{12} - 2 = \frac{5}{2}$

Simplify each term.

Simplify the numerator.

Factor $k$ out of $16 k \cdot 4 + 5 k \cdot 3$ .

Factor $k$ out of $16 k \cdot 4$ .

$\frac{k (16 \cdot 4) + 5 k \cdot 3}{12} - 2 = \frac{5}{2}$

Factor $k$ out of $5 k \cdot 3$ .

$\frac{k (16 \cdot 4) + k (5 \cdot 3)}{12} - 2 = \frac{5}{2}$

Factor $k$ out of $k (16 \cdot 4) + k (5 \cdot 3)$ .

$\frac{k (16 \cdot 4 + 5 \cdot 3)}{12} - 2 = \frac{5}{2}$

Multiply $16$ by $4$ .

$\frac{k (64 + 5 \cdot 3)}{12} - 2 = \frac{5}{2}$

Multiply $5$ by $3$ .

$\frac{k (64 + 15)}{12} - 2 = \frac{5}{2}$

Add $64$ and $15$ .

$\frac{k \cdot 79}{12} - 2 = \frac{5}{2}$

Move $79$ to the left of $k$ .

$\frac{79 k}{12} - 2 = \frac{5}{2}$

Step 5

Move all terms not containing $k$ to the right side of the equation.

Add $2$ to both sides of the equation.

$\frac{79 k}{12} = \frac{5}{2} + 2$

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{79 k}{12} = \frac{5}{2} + 2 \cdot \frac{2}{2}$

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

$\frac{79 k}{12} = \frac{5}{2} + \frac{2 \cdot 2}{2}$

Combine the numerators over the common denominator.

$\frac{79 k}{12} = \frac{5 + 2 \cdot 2}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$\frac{79 k}{12} = \frac{5 + 4}{2}$

Add $5$ and $4$ .

$\frac{79 k}{12} = \frac{9}{2}$

Step 6

Multiply both sides of the equation by $\frac{12}{79}$ .

$\frac{12}{79} \cdot \frac{79 k}{12} = \frac{12}{79} \cdot \frac{9}{2}$

Step 7

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{12}{79} \cdot \frac{79 k}{12}$ .

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12}{79} \cdot \frac{79 k}{12} = \frac{12}{79} \cdot \frac{9}{2}$

Rewrite the expression.

$\frac{1}{79} (79 k) = \frac{12}{79} \cdot \frac{9}{2}$

Cancel the common factor of $79$ .

Factor $79$ out of $79 k$ .

$\frac{1}{79} (79 (k)) = \frac{12}{79} \cdot \frac{9}{2}$

Cancel the common factor.

$\frac{1}{79} (79 k) = \frac{12}{79} \cdot \frac{9}{2}$

Rewrite the expression.

$k = \frac{12}{79} \cdot \frac{9}{2}$

Simplify the right side.

Simplify $\frac{12}{79} \cdot \frac{9}{2}$ .

Cancel the common factor of $2$ .

Factor $2$ out of $12$ .

$k = \frac{2 (6)}{79} \cdot \frac{9}{2}$

Cancel the common factor.

$k = \frac{2 \cdot 6}{79} \cdot \frac{9}{2}$

Rewrite the expression.

$k = \frac{6}{79} \cdot 9$

Combine $\frac{6}{79}$ and $9$ .

$k = \frac{6 \cdot 9}{79}$

Multiply $6$ by $9$ .

$k = \frac{54}{79}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$k = \frac{54}{79}$

Decimal Form:

$k = 0.68354430 \dots$

Solve for k 5/4(2-k)=2(3k-1)-2/3k

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