Solve for x d(-3+x)=kx+9

$d (- 3 + x) = k x + 9$

Step 1

Simplify $d (- 3 + x)$ .

Rewrite.

$0 + 0 + d (- 3 + x) = k x + 9$

Simplify by adding zeros.

$d (- 3 + x) = k x + 9$

Apply the distributive property.

$d \cdot - 3 + d x = k x + 9$

Move $- 3$ to the left of $d$ .

$- 3 d + d x = k x + 9$

Step 2

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

$- 3 d + d x - k x = 9$

Step 3

Add $3 d$ to both sides of the equation.

$d x - k x = 9 + 3 d$

Step 4

Factor $x$ out of $d x - k x$ .

Factor $x$ out of $d x$ .

$x d - k x = 9 + 3 d$

Factor $x$ out of $- k x$ .

$x d + x (- k) = 9 + 3 d$

Factor $x$ out of $x d + x (- k)$ .

$x (d - k) = 9 + 3 d$

Step 5

Divide each term in $x (d - k) = 9 + 3 d$ by $d - k$ and simplify.

Divide each term in $x (d - k) = 9 + 3 d$ by $d - k$ .

$\frac{x (d - k)}{d - k} = \frac{9}{d - k} + \frac{3 d}{d - k}$

Simplify the left side.

Cancel the common factor of $d - k$ .

Cancel the common factor.

$\frac{x (d - k)}{d - k} = \frac{9}{d - k} + \frac{3 d}{d - k}$

Divide $x$ by $1$ .

$x = \frac{9}{d - k} + \frac{3 d}{d - k}$

Simplify the right side.

Combine the numerators over the common denominator.

$x = \frac{9 + 3 d}{d - k}$

Factor $3$ out of $9 + 3 d$ .

Factor $3$ out of $9$ .

$x = \frac{3 \cdot 3 + 3 d}{d - k}$

Factor $3$ out of $3 \cdot 3 + 3 d$ .

$x = \frac{3 (3 + d)}{d - k}$