Solve p=1/3+4/3a/frac{1{3a}+4/3a^2}

$$p = \frac { \frac { 1 } { 3 } + \frac { 4 } { 3 a } } { \frac { 1 } { 3 a } + \frac { 4 } { 3 a ^ { 2 } } }$$

Solve for p

$$p=a$$

$$a\neq 0\text{ and }a\neq -4$$

Solution Steps

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

$$p=\frac{\frac{a}{3a}+\frac{4}{3a}}{\frac{1}{3a}+\frac{4}{3a^{2}}}$$

Since \frac{a}{3a} and \frac{4}{3a} have the same denominator, add them by adding their numerators.

$$p=\frac{\frac{a+4}{3a}}{\frac{1}{3a}+\frac{4}{3a^{2}}}$$

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

$$p=\frac{\frac{a+4}{3a}}{\frac{a}{3a^{2}}+\frac{4}{3a^{2}}}$$

Since \frac{a}{3a^{2}} and \frac{4}{3a^{2}} have the same denominator, add them by adding their numerators.

$$p=\frac{\frac{a+4}{3a}}{\frac{a+4}{3a^{2}}}$$

Divide \frac{a+4}{3a} by \frac{a+4}{3a^{2}} by multiplying \frac{a+4}{3a} by the reciprocal of \frac{a+4}{3a^{2}}.

$$p=\frac{\left(a+4\right)\times 3a^{2}}{3a\left(a+4\right)}$$

Cancel out 3a\left(a+4\right) in both numerator and denominator.

$$p=a$$

Solve for a

$$a=p$$

$$p\neq 0\text{ and }p\neq -4$$