Solve for k 2(2k-1)=4(k-2)

$2 (2 k - 1) = 4 (k - 2)$

Step 1

Simplify $2 (2 k - 1)$ .

Start simplifying.

$0 + 0 + 2 (2 k - 1) = 4 (k - 2)$

Simplify by adding zeros.

$2 (2 k - 1) = 4 (k - 2)$

Apply the distributive property.

$2 (2 k) + 2 \cdot - 1 = 4 (k - 2)$

Multiply.

Multiply $2$ by $2$ .

$4 k + 2 \cdot - 1 = 4 (k - 2)$

Multiply $2$ by $- 1$ .

$4 k - 2 = 4 (k - 2)$

Step 2

Simplify $4 (k - 2)$ .

Apply the distributive property.

$4 k - 2 = 4 k + 4 \cdot - 2$

Multiply $4$ by $- 2$ .

$4 k - 2 = 4 k - 8$

Step 3

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

Subtract $4 k$ from both sides of the equation.

$4 k - 2 - 4 k = - 8$

Combine the opposite terms in $4 k - 2 - 4 k$ .

Subtract $4 k$ from $4 k$ .

$0 - 2 = - 8$

Subtract $2$ from $0$ .

$- 2 = - 8$

Step 4

Since $- 2 \neq - 8$ , there are no solutions.

No solution