Solve c+6b/ac+2bc-6ab-3a^2+2b/a^2+2ab-b/ac-3a^2=

$$\frac { c + 6 b } { a c + 2 b c - 6 a b - 3 a ^ { 2 } } + \frac { 2 b } { a ^ { 2 } + 2 a b } - \frac { b } { a c - 3 a ^ { 2 } } =$$

Evaluate

$$-\frac{c-b}{a\left(3a-c\right)}$$

Short Solution Steps

Factor ac+2bc-6ab-3a^{2}. Factor a^{2}+2ab.

$$\frac{c+6b}{\left(-a-2b\right)\left(3a-c\right)}+\frac{2b}{a\left(a+2b\right)}-\frac{b}{ac-3a^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(-a-2b\right)\left(3a-c\right) and a\left(a+2b\right) is a\left(-3a+c\right)\left(-a-2b\right). Multiply \frac{c+6b}{\left(-a-2b\right)\left(3a-c\right)} times \frac{-a}{-a}. Multiply \frac{2b}{a\left(a+2b\right)} times \frac{-\left(-3a+c\right)}{-\left(-3a+c\right)}.

$$\frac{\left(c+6b\right)\left(-1\right)a}{a\left(-3a+c\right)\left(-a-2b\right)}+\frac{2b\left(-1\right)\left(-3a+c\right)}{a\left(-3a+c\right)\left(-a-2b\right)}-\frac{b}{ac-3a^{2}}$$

Since \frac{\left(c+6b\right)\left(-1\right)a}{a\left(-3a+c\right)\left(-a-2b\right)} and \frac{2b\left(-1\right)\left(-3a+c\right)}{a\left(-3a+c\right)\left(-a-2b\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(c+6b\right)\left(-1\right)a+2b\left(-1\right)\left(-3a+c\right)}{a\left(-3a+c\right)\left(-a-2b\right)}-\frac{b}{ac-3a^{2}}$$

Do the multiplications in \left(c+6b\right)\left(-1\right)a+2b\left(-1\right)\left(-3a+c\right).

$$\frac{-ca-6ba+6ba-2bc}{a\left(-3a+c\right)\left(-a-2b\right)}-\frac{b}{ac-3a^{2}}$$

Combine like terms in -ca-6ba+6ba-2bc.

$$\frac{-ca-2bc}{a\left(-3a+c\right)\left(-a-2b\right)}-\frac{b}{ac-3a^{2}}$$

Factor the expressions that are not already factored in \frac{-ca-2bc}{a\left(-3a+c\right)\left(-a-2b\right)}.

$$\frac{c\left(-a-2b\right)}{a\left(-3a+c\right)\left(-a-2b\right)}-\frac{b}{ac-3a^{2}}$$

Cancel out -a-2b in both numerator and denominator.

$$\frac{c}{a\left(-3a+c\right)}-\frac{b}{ac-3a^{2}}$$

Factor ac-3a^{2}.

$$\frac{c}{a\left(-3a+c\right)}-\frac{b}{a\left(-3a+c\right)}$$

Since \frac{c}{a\left(-3a+c\right)} and \frac{b}{a\left(-3a+c\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{c-b}{a\left(-3a+c\right)}$$

Expand a\left(-3a+c\right).

$$\frac{c-b}{-3a^{2}+ac}$$

Expand

$$\frac{b-c}{a\left(3a-c\right)}$$

Short Solution Steps

Factor ac+2bc-6ab-3a^{2}. Factor a^{2}+2ab.

$$\frac{c+6b}{\left(-a-2b\right)\left(3a-c\right)}+\frac{2b}{a\left(a+2b\right)}-\frac{b}{ac-3a^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(-a-2b\right)\left(3a-c\right) and a\left(a+2b\right) is a\left(-3a+c\right)\left(-a-2b\right). Multiply \frac{c+6b}{\left(-a-2b\right)\left(3a-c\right)} times \frac{-a}{-a}. Multiply \frac{2b}{a\left(a+2b\right)} times \frac{-\left(-3a+c\right)}{-\left(-3a+c\right)}.

$$\frac{\left(c+6b\right)\left(-1\right)a}{a\left(-3a+c\right)\left(-a-2b\right)}+\frac{2b\left(-1\right)\left(-3a+c\right)}{a\left(-3a+c\right)\left(-a-2b\right)}-\frac{b}{ac-3a^{2}}$$

Since \frac{\left(c+6b\right)\left(-1\right)a}{a\left(-3a+c\right)\left(-a-2b\right)} and \frac{2b\left(-1\right)\left(-3a+c\right)}{a\left(-3a+c\right)\left(-a-2b\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(c+6b\right)\left(-1\right)a+2b\left(-1\right)\left(-3a+c\right)}{a\left(-3a+c\right)\left(-a-2b\right)}-\frac{b}{ac-3a^{2}}$$

Do the multiplications in \left(c+6b\right)\left(-1\right)a+2b\left(-1\right)\left(-3a+c\right).

$$\frac{-ca-6ba+6ba-2bc}{a\left(-3a+c\right)\left(-a-2b\right)}-\frac{b}{ac-3a^{2}}$$

Combine like terms in -ca-6ba+6ba-2bc.

$$\frac{-ca-2bc}{a\left(-3a+c\right)\left(-a-2b\right)}-\frac{b}{ac-3a^{2}}$$

Factor the expressions that are not already factored in \frac{-ca-2bc}{a\left(-3a+c\right)\left(-a-2b\right)}.

$$\frac{c\left(-a-2b\right)}{a\left(-3a+c\right)\left(-a-2b\right)}-\frac{b}{ac-3a^{2}}$$

Cancel out -a-2b in both numerator and denominator.

$$\frac{c}{a\left(-3a+c\right)}-\frac{b}{ac-3a^{2}}$$

Factor ac-3a^{2}.

$$\frac{c}{a\left(-3a+c\right)}-\frac{b}{a\left(-3a+c\right)}$$

Since \frac{c}{a\left(-3a+c\right)} and \frac{b}{a\left(-3a+c\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{c-b}{a\left(-3a+c\right)}$$

Expand a\left(-3a+c\right).

$$\frac{c-b}{-3a^{2}+ac}$$