Solve a-3/3a^2+2-a/a

$$\frac{ a-3 }{ 3 { a }^{ 2 } } + \frac{ 2-a }{ a }$$

Evaluate

$$-1+\frac{7}{3a}-\frac{1}{a^{2}}$$

Short Solution Steps

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

$$\frac{a-3}{3a^{2}}+\frac{\left(2-a\right)\times 3a}{3a^{2}}$$

Since \frac{a-3}{3a^{2}} and \frac{\left(2-a\right)\times 3a}{3a^{2}} have the same denominator, add them by adding their numerators.

$$\frac{a-3+\left(2-a\right)\times 3a}{3a^{2}}$$

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

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

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

$$\frac{7a-3-3a^{2}}{3a^{2}}$$

Factor the expressions that are not already factored in \frac{7a-3-3a^{2}}{3a^{2}}.

$$\frac{-3\left(a-\left(-\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\right)\left(a-\left(\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\right)}{3a^{2}}$$

Cancel out 3 in both numerator and denominator.

$$\frac{-\left(a-\left(-\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\right)\left(a-\left(\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\right)}{a^{2}}$$

To find the opposite of -\frac{1}{6}\sqrt{13}+\frac{7}{6}, find the opposite of each term.

$$\frac{-\left(a+\frac{1}{6}\sqrt{13}-\frac{7}{6}\right)\left(a-\left(\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\right)}{a^{2}}$$

To find the opposite of \frac{1}{6}\sqrt{13}+\frac{7}{6}, find the opposite of each term.

$$\frac{-\left(a+\frac{1}{6}\sqrt{13}-\frac{7}{6}\right)\left(a-\frac{1}{6}\sqrt{13}-\frac{7}{6}\right)}{a^{2}}$$

Use the distributive property to multiply -1 by a+\frac{1}{6}\sqrt{13}-\frac{7}{6}.

$$\frac{\left(-a-\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\left(a-\frac{1}{6}\sqrt{13}-\frac{7}{6}\right)}{a^{2}}$$

Use the distributive property to multiply -a-\frac{1}{6}\sqrt{13}+\frac{7}{6} by a-\frac{1}{6}\sqrt{13}-\frac{7}{6} and combine like terms.

$$\frac{-a^{2}+\frac{7}{3}a+\frac{1}{36}\left(\sqrt{13}\right)^{2}-\frac{49}{36}}{a^{2}}$$

The square of \sqrt{13} is 13.

$$\frac{-a^{2}+\frac{7}{3}a+\frac{1}{36}\times 13-\frac{49}{36}}{a^{2}}$$

Multiply \frac{1}{36} and 13 to get \frac{13}{36}.

$$\frac{-a^{2}+\frac{7}{3}a+\frac{13}{36}-\frac{49}{36}}{a^{2}}$$

Subtract \frac{49}{36} from \frac{13}{36} to get -1.

$$\frac{-a^{2}+\frac{7}{3}a-1}{a^{2}}$$

Expand

$$-1+\frac{7}{3a}-\frac{1}{a^{2}}$$

Short Solution Steps

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

$$\frac{a-3}{3a^{2}}+\frac{\left(2-a\right)\times 3a}{3a^{2}}$$

Since \frac{a-3}{3a^{2}} and \frac{\left(2-a\right)\times 3a}{3a^{2}} have the same denominator, add them by adding their numerators.

$$\frac{a-3+\left(2-a\right)\times 3a}{3a^{2}}$$

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

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

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

$$\frac{7a-3-3a^{2}}{3a^{2}}$$

Factor the expressions that are not already factored in \frac{7a-3-3a^{2}}{3a^{2}}.

$$\frac{-3\left(a-\left(-\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\right)\left(a-\left(\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\right)}{3a^{2}}$$

Cancel out 3 in both numerator and denominator.

$$\frac{-\left(a-\left(-\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\right)\left(a-\left(\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\right)}{a^{2}}$$

To find the opposite of -\frac{1}{6}\sqrt{13}+\frac{7}{6}, find the opposite of each term.

$$\frac{-\left(a+\frac{1}{6}\sqrt{13}-\frac{7}{6}\right)\left(a-\left(\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\right)}{a^{2}}$$

To find the opposite of \frac{1}{6}\sqrt{13}+\frac{7}{6}, find the opposite of each term.

$$\frac{-\left(a+\frac{1}{6}\sqrt{13}-\frac{7}{6}\right)\left(a-\frac{1}{6}\sqrt{13}-\frac{7}{6}\right)}{a^{2}}$$

Use the distributive property to multiply -1 by a+\frac{1}{6}\sqrt{13}-\frac{7}{6}.

$$\frac{\left(-a-\frac{1}{6}\sqrt{13}+\frac{7}{6}\right)\left(a-\frac{1}{6}\sqrt{13}-\frac{7}{6}\right)}{a^{2}}$$

Use the distributive property to multiply -a-\frac{1}{6}\sqrt{13}+\frac{7}{6} by a-\frac{1}{6}\sqrt{13}-\frac{7}{6} and combine like terms.

$$\frac{-a^{2}+\frac{7}{3}a+\frac{1}{36}\left(\sqrt{13}\right)^{2}-\frac{49}{36}}{a^{2}}$$

The square of \sqrt{13} is 13.

$$\frac{-a^{2}+\frac{7}{3}a+\frac{1}{36}\times 13-\frac{49}{36}}{a^{2}}$$

Multiply \frac{1}{36} and 13 to get \frac{13}{36}.

$$\frac{-a^{2}+\frac{7}{3}a+\frac{13}{36}-\frac{49}{36}}{a^{2}}$$

Subtract \frac{49}{36} from \frac{13}{36} to get -1.

$$\frac{-a^{2}+\frac{7}{3}a-1}{a^{2}}$$