Solve (2/3a-1/3b+1)*(1/2a-3/4b)-(frac{2}{3}a^4-frac{1}{3}a^3b-frac{1}{2}a^2b):(frac{2}{3}a^2)

$$( \frac { 2 } { 3 } a - \frac { 1 } { 3 } b + 1 ) \cdot ( \frac { 1 } { 2 } a - \frac { 3 } { 4 } b ) - ( \frac { 2 } { 3 } a ^ { 4 } - \frac { 1 } { 3 } a ^ { 3 } b - \frac { 1 } { 2 } a ^ { 2 } b ) : ( \frac { 2 } { 3 } a ^ { 2 } )$$

Evaluate

$$-\frac{ab}{6}+\frac{b^{2}}{4}-\frac{2a^{2}}{3}+\frac{a}{2}$$

Short Solution Steps

Use the distributive property to multiply \frac{2}{3}a-\frac{1}{3}b+1 by \frac{1}{2}a-\frac{3}{4}b and combine like terms.

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-\frac{\frac{2}{3}a^{4}-\frac{1}{3}a^{3}b-\frac{1}{2}a^{2}b}{\frac{2}{3}a^{2}}$$

Factor the expressions that are not already factored in \frac{\frac{2}{3}a^{4}-\frac{1}{3}a^{3}b-\frac{1}{2}a^{2}b}{\frac{2}{3}a^{2}}.

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-\frac{\frac{1}{6}a^{2}\left(4a^{2}-2ab-3b\right)}{\frac{2}{3}a^{2}}$$

Cancel out a^{2} in both numerator and denominator.

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-\frac{\frac{1}{6}\left(4a^{2}-2ab-3b\right)}{\frac{2}{3}}$$

Divide \frac{1}{6}\left(4a^{2}-2ab-3b\right) by \frac{2}{3} to get \frac{1}{4}\left(4a^{2}-2ab-3b\right).

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-\frac{1}{4}\left(4a^{2}-2ab-3b\right)$$

Use the distributive property to multiply \frac{1}{4} by 4a^{2}-2ab-3b.

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-\left(a^{2}-\frac{1}{2}ab-\frac{3}{4}b\right)$$

To find the opposite of a^{2}-\frac{1}{2}ab-\frac{3}{4}b, find the opposite of each term.

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-a^{2}+\frac{1}{2}ab+\frac{3}{4}b$$

Combine \frac{1}{3}a^{2} and -a^{2} to get -\frac{2}{3}a^{2}.

$$-\frac{2}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b+\frac{1}{2}ab+\frac{3}{4}b$$

Combine -\frac{2}{3}ab and \frac{1}{2}ab to get -\frac{1}{6}ab.

$$-\frac{2}{3}a^{2}-\frac{1}{6}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b+\frac{3}{4}b$$

Combine -\frac{3}{4}b and \frac{3}{4}b to get 0.

$$-\frac{2}{3}a^{2}-\frac{1}{6}ab+\frac{1}{4}b^{2}+\frac{1}{2}a$$

Expand

$$-\frac{ab}{6}+\frac{b^{2}}{4}-\frac{2a^{2}}{3}+\frac{a}{2}$$

Short Solution Steps

Use the distributive property to multiply \frac{2}{3}a-\frac{1}{3}b+1 by \frac{1}{2}a-\frac{3}{4}b and combine like terms.

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-\frac{\frac{2}{3}a^{4}-\frac{1}{3}a^{3}b-\frac{1}{2}a^{2}b}{\frac{2}{3}a^{2}}$$

Factor the expressions that are not already factored in \frac{\frac{2}{3}a^{4}-\frac{1}{3}a^{3}b-\frac{1}{2}a^{2}b}{\frac{2}{3}a^{2}}.

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-\frac{\frac{1}{6}a^{2}\left(4a^{2}-2ab-3b\right)}{\frac{2}{3}a^{2}}$$

Cancel out a^{2} in both numerator and denominator.

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-\frac{\frac{1}{6}\left(4a^{2}-2ab-3b\right)}{\frac{2}{3}}$$

Divide \frac{1}{6}\left(4a^{2}-2ab-3b\right) by \frac{2}{3} to get \frac{1}{4}\left(4a^{2}-2ab-3b\right).

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-\frac{1}{4}\left(4a^{2}-2ab-3b\right)$$

Use the distributive property to multiply \frac{1}{4} by 4a^{2}-2ab-3b.

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-\left(a^{2}-\frac{1}{2}ab-\frac{3}{4}b\right)$$

To find the opposite of a^{2}-\frac{1}{2}ab-\frac{3}{4}b, find the opposite of each term.

$$\frac{1}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b-a^{2}+\frac{1}{2}ab+\frac{3}{4}b$$

Combine \frac{1}{3}a^{2} and -a^{2} to get -\frac{2}{3}a^{2}.

$$-\frac{2}{3}a^{2}-\frac{2}{3}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b+\frac{1}{2}ab+\frac{3}{4}b$$

Combine -\frac{2}{3}ab and \frac{1}{2}ab to get -\frac{1}{6}ab.

$$-\frac{2}{3}a^{2}-\frac{1}{6}ab+\frac{1}{4}b^{2}+\frac{1}{2}a-\frac{3}{4}b+\frac{3}{4}b$$

Combine -\frac{3}{4}b and \frac{3}{4}b to get 0.

$$-\frac{2}{3}a^{2}-\frac{1}{6}ab+\frac{1}{4}b^{2}+\frac{1}{2}a$$