Solve the Inequality for x b(x-3)>=ax+7b

$b (x - 3) \geq a x + 7 b$

Simplify $b (x - 3)$ .

Rewrite.

$0 + 0 + b (x - 3) \geq a x + 7 b$

Simplify by adding zeros.

$b (x - 3) \geq a x + 7 b$

Apply the distributive property.

$b x + b \cdot - 3 \geq a x + 7 b$

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

$b x - 3 b \geq a x + 7 b$

Subtract $a x$ from both sides of the inequality.

$b x - 3 b - a x \geq 7 b$

Move all terms not containing $x$ to the right side of the inequality.

Add $3 b$ to both sides of the inequality.

$b x - a x \geq 7 b + 3 b$

Add $7 b$ and $3 b$ .

$b x - a x \geq 10 b$

Factor $x$ out of $b x - a x$ .

Factor $x$ out of $b x$ .

$x b - a x \geq 10 b$

Factor $x$ out of $- a x$ .

$x b + x (- a) \geq 10 b$

Factor $x$ out of $x b + x (- a)$ .

$x (b - a) \geq 10 b$

Divide each term in $x (b - a) \geq 10 b$ by $b - a$ and simplify.

Divide each term in $x (b - a) \geq 10 b$ by $b - a$ .

$\frac{x (b - a)}{b - a} \geq \frac{10 b}{b - a}$

Simplify the left side.

Cancel the common factor of $b - a$ .

Cancel the common factor.

$\frac{x (b - a)}{b - a} \geq \frac{10 b}{b - a}$

Divide $x$ by $1$ .

$x \geq \frac{10 b}{b - a}$