Solve (9/6bc-1/3b^2)(-2/3c)-[2/9b^2c-(+frac{1}{5}bc)*(frac{15}{3}c-frac{25}{2}b)]

$$( \frac { 9 } { 6 } b c - \frac { 1 } { 3 } b ^ { 2 } ) ( - \frac { 2 } { 3 } c ) - [ \frac { 2 } { 9 } b ^ { 2 } c - ( + \frac { 1 } { 5 } b c ) \cdot ( \frac { 15 } { 3 } c - \frac { 25 } { 2 } b ) ]$$

Evaluate

$$-\frac{5cb^{2}}{2}$$

Solution Steps

Reduce the fraction \frac{9}{6} to lowest terms by extracting and canceling out 3.

$$\left(\frac{3}{2}bc-\frac{1}{3}b^{2}\right)\left(-\frac{2}{3}\right)c-\left(\frac{2}{9}b^{2}c-\frac{1}{5}bc\left(\frac{15}{3}c-\frac{25}{2}b\right)\right)$$

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

$$\left(-bc+\frac{2}{9}b^{2}\right)c-\left(\frac{2}{9}b^{2}c-\frac{1}{5}bc\left(\frac{15}{3}c-\frac{25}{2}b\right)\right)$$

Use the distributive property to multiply -bc+\frac{2}{9}b^{2} by c.

$$-bc^{2}+\frac{2}{9}b^{2}c-\left(\frac{2}{9}b^{2}c-\frac{1}{5}bc\left(\frac{15}{3}c-\frac{25}{2}b\right)\right)$$

Divide 15 by 3 to get 5.

$$-bc^{2}+\frac{2}{9}b^{2}c-\left(\frac{2}{9}b^{2}c-\frac{1}{5}bc\left(5c-\frac{25}{2}b\right)\right)$$

Use the distributive property to multiply -\frac{1}{5}bc by 5c-\frac{25}{2}b.

$$-bc^{2}+\frac{2}{9}b^{2}c-\left(\frac{2}{9}b^{2}c-bc^{2}+\frac{5}{2}b^{2}c\right)$$

Combine \frac{2}{9}b^{2}c and \frac{5}{2}b^{2}c to get \frac{49}{18}b^{2}c.

$$-bc^{2}+\frac{2}{9}b^{2}c-\left(\frac{49}{18}b^{2}c-bc^{2}\right)$$

To find the opposite of \frac{49}{18}b^{2}c-bc^{2}, find the opposite of each term.

$$-bc^{2}+\frac{2}{9}b^{2}c-\frac{49}{18}b^{2}c+bc^{2}$$

Combine \frac{2}{9}b^{2}c and -\frac{49}{18}b^{2}c to get -\frac{5}{2}b^{2}c.

$$-bc^{2}-\frac{5}{2}b^{2}c+bc^{2}$$

Combine -bc^{2} and bc^{2} to get 0.

$$-\frac{5}{2}b^{2}c$$

Expand

$$-\frac{5cb^{2}}{2}$$

Solution Steps

Reduce the fraction \frac{9}{6} to lowest terms by extracting and canceling out 3.

$$\left(\frac{3}{2}bc-\frac{1}{3}b^{2}\right)\left(-\frac{2}{3}\right)c-\left(\frac{2}{9}b^{2}c-\frac{1}{5}bc\left(\frac{15}{3}c-\frac{25}{2}b\right)\right)$$

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

$$\left(-bc+\frac{2}{9}b^{2}\right)c-\left(\frac{2}{9}b^{2}c-\frac{1}{5}bc\left(\frac{15}{3}c-\frac{25}{2}b\right)\right)$$

Use the distributive property to multiply -bc+\frac{2}{9}b^{2} by c.

$$-bc^{2}+\frac{2}{9}b^{2}c-\left(\frac{2}{9}b^{2}c-\frac{1}{5}bc\left(\frac{15}{3}c-\frac{25}{2}b\right)\right)$$

Divide 15 by 3 to get 5.

$$-bc^{2}+\frac{2}{9}b^{2}c-\left(\frac{2}{9}b^{2}c-\frac{1}{5}bc\left(5c-\frac{25}{2}b\right)\right)$$

Use the distributive property to multiply -\frac{1}{5}bc by 5c-\frac{25}{2}b.

$$-bc^{2}+\frac{2}{9}b^{2}c-\left(\frac{2}{9}b^{2}c-bc^{2}+\frac{5}{2}b^{2}c\right)$$

Combine \frac{2}{9}b^{2}c and \frac{5}{2}b^{2}c to get \frac{49}{18}b^{2}c.

$$-bc^{2}+\frac{2}{9}b^{2}c-\left(\frac{49}{18}b^{2}c-bc^{2}\right)$$

To find the opposite of \frac{49}{18}b^{2}c-bc^{2}, find the opposite of each term.

$$-bc^{2}+\frac{2}{9}b^{2}c-\frac{49}{18}b^{2}c+bc^{2}$$

Combine \frac{2}{9}b^{2}c and -\frac{49}{18}b^{2}c to get -\frac{5}{2}b^{2}c.

$$-bc^{2}-\frac{5}{2}b^{2}c+bc^{2}$$

Combine -bc^{2} and bc^{2} to get 0.

$$-\frac{5}{2}b^{2}c$$