Solve for k 3(2k-5)=6(k-4)+9

$3 (2 k - 5) = 6 (k - 4) + 9$

Simplify $3 (2 k - 5)$ .

Rewrite.

$0 + 0 + 3 (2 k - 5) = 6 (k - 4) + 9$

Simplify by adding zeros.

$3 (2 k - 5) = 6 (k - 4) + 9$

Apply the distributive property.

$3 (2 k) + 3 \cdot - 5 = 6 (k - 4) + 9$

Multiply.

Multiply $2$ by $3$ .

$6 k + 3 \cdot - 5 = 6 (k - 4) + 9$

Multiply $3$ by $- 5$ .

$6 k - 15 = 6 (k - 4) + 9$

Simplify $6 (k - 4) + 9$ .

Simplify each term.

Apply the distributive property.

$6 k - 15 = 6 k + 6 \cdot - 4 + 9$

Multiply $6$ by $- 4$ .

$6 k - 15 = 6 k - 24 + 9$

Add $- 24$ and $9$ .

$6 k - 15 = 6 k - 15$

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

Subtract $6 k$ from both sides of the equation.

$6 k - 15 - 6 k = - 15$

Combine the opposite terms in $6 k - 15 - 6 k$ .

Subtract $6 k$ from $6 k$ .

$0 - 15 = - 15$

Subtract $15$ from $0$ .

$- 15 = - 15$

Since $- 15 = - 15$ , the equation will always be true for any value of $k$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$