Solve (3a/1-3a-2a/3a+1):6a^2+10a/1-6a+9a^2

$$( \frac { 3 a } { 1 - 3 a } - \frac { 2 a } { 3 a + 1 } ) : \frac { 6 a ^ { 2 } + 10 a } { 1 - 6 a + 9 a ^ { 2 } }$$

Evaluate

$$-\frac{\left(3a-1\right)\left(15a+1\right)}{2\left(3a+1\right)\left(3a+5\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 1-3a and 3a+1 is \left(-3a+1\right)\left(3a+1\right). Multiply \frac{3a}{1-3a} times \frac{3a+1}{3a+1}. Multiply \frac{2a}{3a+1} times \frac{-3a+1}{-3a+1}.

$$\frac{\frac{3a\left(3a+1\right)}{\left(-3a+1\right)\left(3a+1\right)}-\frac{2a\left(-3a+1\right)}{\left(-3a+1\right)\left(3a+1\right)}}{\frac{6a^{2}+10a}{1-6a+9a^{2}}}$$

Since \frac{3a\left(3a+1\right)}{\left(-3a+1\right)\left(3a+1\right)} and \frac{2a\left(-3a+1\right)}{\left(-3a+1\right)\left(3a+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3a\left(3a+1\right)-2a\left(-3a+1\right)}{\left(-3a+1\right)\left(3a+1\right)}}{\frac{6a^{2}+10a}{1-6a+9a^{2}}}$$

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

$$\frac{\frac{9a^{2}+3a+6a^{2}-2a}{\left(-3a+1\right)\left(3a+1\right)}}{\frac{6a^{2}+10a}{1-6a+9a^{2}}}$$

Combine like terms in 9a^{2}+3a+6a^{2}-2a.

$$\frac{\frac{15a^{2}+a}{\left(-3a+1\right)\left(3a+1\right)}}{\frac{6a^{2}+10a}{1-6a+9a^{2}}}$$

Divide \frac{15a^{2}+a}{\left(-3a+1\right)\left(3a+1\right)} by \frac{6a^{2}+10a}{1-6a+9a^{2}} by multiplying \frac{15a^{2}+a}{\left(-3a+1\right)\left(3a+1\right)} by the reciprocal of \frac{6a^{2}+10a}{1-6a+9a^{2}}.

$$\frac{\left(15a^{2}+a\right)\left(1-6a+9a^{2}\right)}{\left(-3a+1\right)\left(3a+1\right)\left(6a^{2}+10a\right)}$$

Factor the expressions that are not already factored.

$$\frac{a\left(15a+1\right)\left(3a-1\right)^{2}}{2a\left(-3a+1\right)\left(3a+1\right)\left(3a+5\right)}$$

Cancel out a in both numerator and denominator.

$$\frac{\left(15a+1\right)\left(3a-1\right)^{2}}{2\left(-3a+1\right)\left(3a+1\right)\left(3a+5\right)}$$

Expand the expression.

$$\frac{135a^{3}-81a^{2}+9a+1}{-54a^{3}-90a^{2}+6a+10}$$

Factor the expressions that are not already factored.

$$\frac{\left(15a+1\right)\left(3a-1\right)^{2}}{2\left(-3a-1\right)\left(3a-1\right)\left(3a+5\right)}$$

Cancel out 3a-1 in both numerator and denominator.

$$\frac{\left(3a-1\right)\left(15a+1\right)}{2\left(-3a-1\right)\left(3a+5\right)}$$

Expand the expression.

$$\frac{45a^{2}-12a-1}{-18a^{2}-36a-10}$$

Expand

$$-\frac{45a^{2}-12a-1}{2\left(3a+1\right)\left(3a+5\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 1-3a and 3a+1 is \left(-3a+1\right)\left(3a+1\right). Multiply \frac{3a}{1-3a} times \frac{3a+1}{3a+1}. Multiply \frac{2a}{3a+1} times \frac{-3a+1}{-3a+1}.

$$\frac{\frac{3a\left(3a+1\right)}{\left(-3a+1\right)\left(3a+1\right)}-\frac{2a\left(-3a+1\right)}{\left(-3a+1\right)\left(3a+1\right)}}{\frac{6a^{2}+10a}{1-6a+9a^{2}}}$$

Since \frac{3a\left(3a+1\right)}{\left(-3a+1\right)\left(3a+1\right)} and \frac{2a\left(-3a+1\right)}{\left(-3a+1\right)\left(3a+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{3a\left(3a+1\right)-2a\left(-3a+1\right)}{\left(-3a+1\right)\left(3a+1\right)}}{\frac{6a^{2}+10a}{1-6a+9a^{2}}}$$

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

$$\frac{\frac{9a^{2}+3a+6a^{2}-2a}{\left(-3a+1\right)\left(3a+1\right)}}{\frac{6a^{2}+10a}{1-6a+9a^{2}}}$$

Combine like terms in 9a^{2}+3a+6a^{2}-2a.

$$\frac{\frac{15a^{2}+a}{\left(-3a+1\right)\left(3a+1\right)}}{\frac{6a^{2}+10a}{1-6a+9a^{2}}}$$

Divide \frac{15a^{2}+a}{\left(-3a+1\right)\left(3a+1\right)} by \frac{6a^{2}+10a}{1-6a+9a^{2}} by multiplying \frac{15a^{2}+a}{\left(-3a+1\right)\left(3a+1\right)} by the reciprocal of \frac{6a^{2}+10a}{1-6a+9a^{2}}.

$$\frac{\left(15a^{2}+a\right)\left(1-6a+9a^{2}\right)}{\left(-3a+1\right)\left(3a+1\right)\left(6a^{2}+10a\right)}$$

Factor the expressions that are not already factored.

$$\frac{a\left(15a+1\right)\left(3a-1\right)^{2}}{2a\left(-3a+1\right)\left(3a+1\right)\left(3a+5\right)}$$

Cancel out a in both numerator and denominator.

$$\frac{\left(15a+1\right)\left(3a-1\right)^{2}}{2\left(-3a+1\right)\left(3a+1\right)\left(3a+5\right)}$$

Expand the expression.

$$\frac{135a^{3}-81a^{2}+9a+1}{-54a^{3}-90a^{2}+6a+10}$$

Factor the expressions that are not already factored.

$$\frac{\left(15a+1\right)\left(3a-1\right)^{2}}{2\left(-3a-1\right)\left(3a-1\right)\left(3a+5\right)}$$

Cancel out 3a-1 in both numerator and denominator.

$$\frac{\left(3a-1\right)\left(15a+1\right)}{2\left(-3a-1\right)\left(3a+5\right)}$$

Expand the expression.

$$\frac{45a^{2}-12a-1}{-18a^{2}-36a-10}$$