Solve the Inequality for x a(bx-c)>=bc

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

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

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

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

Simplify the left side.

Cancel the common factor of $a$ .

Cancel the common factor.

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

Divide $b x - c$ by $1$ .

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

Add $c$ to both sides of the inequality.

$b x \geq \frac{b c}{a} + c$

Divide each term in $b x \geq \frac{b c}{a} + c$ by $b$ and simplify.

Divide each term in $b x \geq \frac{b c}{a} + c$ by $b$ .

$\frac{b x}{b} \geq \frac{\frac{b c}{a}}{b} + \frac{c}{b}$

Simplify the left side.

Cancel the common factor of $b$ .

Cancel the common factor.

$\frac{b x}{b} \geq \frac{\frac{b c}{a}}{b} + \frac{c}{b}$

Divide $x$ by $1$ .

$x \geq \frac{\frac{b c}{a}}{b} + \frac{c}{b}$

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$x \geq \frac{b c}{a} \cdot \frac{1}{b} + \frac{c}{b}$

Cancel the common factor of $b$ .

Factor $b$ out of $b c$ .

$x \geq \frac{b (c)}{a} \cdot \frac{1}{b} + \frac{c}{b}$

Cancel the common factor.

$x \geq \frac{b c}{a} \cdot \frac{1}{b} + \frac{c}{b}$

Rewrite the expression.

$x \geq \frac{c}{a} + \frac{c}{b}$