Solve for k 3(4k-1/3)=12k-1

$3 (4 k - \frac{1}{3}) = 12 k - 1$

Step 1

Simplify $3 (4 k - \frac{1}{3})$ .

Rewrite.

$0 + 0 + 3 (4 k - \frac{1}{3}) = 12 k - 1$

Simplify by adding zeros.

$3 (4 k - \frac{1}{3}) = 12 k - 1$

Apply the distributive property.

$3 (4 k) + 3 (- \frac{1}{3}) = 12 k - 1$

Multiply $4$ by $3$ .

$12 k + 3 (- \frac{1}{3}) = 12 k - 1$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$12 k + 3 (\frac{- 1}{3}) = 12 k - 1$

Cancel the common factor.

$12 k + 3 (\frac{- 1}{3}) = 12 k - 1$

Rewrite the expression.

$12 k - 1 = 12 k - 1$

Step 2

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

Subtract $12 k$ from both sides of the equation.

$12 k - 1 - 12 k = - 1$

Combine the opposite terms in $12 k - 1 - 12 k$ .

Subtract $12 k$ from $12 k$ .

$0 - 1 = - 1$

Subtract $1$ from $0$ .

$- 1 = - 1$

Step 3

Since $- 1 = - 1$ , the equation will always be true for any value of $k$ .

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$