Solve (9/2a^6bc+5/6a^7b):(-1/6a^5b)

$$( \frac { 9 } { 2 } a ^ { 6 } b c + \frac { 5 } { 6 } a ^ { 7 } b ) : ( - \frac { 1 } { 6 } a ^ { 5 } b )$$

Evaluate

$$-a\left(5a+27c\right)$$

Short Solution Steps

Factor the expressions that are not already factored.

$$\frac{\frac{1}{6}b\left(5a+27c\right)a^{6}}{-\frac{1}{6}ba^{5}}$$

Cancel out ba^{5} in both numerator and denominator.

$$\frac{\frac{1}{6}a\left(5a+27c\right)}{-\frac{1}{6}}$$

Expand the expression.

$$\frac{\frac{5}{6}a^{2}+\frac{9}{2}ac}{-\frac{1}{6}}$$

Divide \frac{5}{6}a^{2}+\frac{9}{2}ac by -\frac{1}{6} by multiplying \frac{5}{6}a^{2}+\frac{9}{2}ac by the reciprocal of -\frac{1}{6}.

$$\frac{\left(\frac{5}{6}a^{2}+\frac{9}{2}ac\right)\times 6}{-1}$$

Anything divided by -1 gives its opposite.

$$-\left(\frac{5}{6}a^{2}+\frac{9}{2}ac\right)\times 6$$

Use the distributive property to multiply \frac{5}{6}a^{2}+\frac{9}{2}ac by 6.

$$-\left(5a^{2}+27ac\right)$$

To find the opposite of 5a^{2}+27ac, find the opposite of each term.

$$-5a^{2}-27ac$$

Expand

$$-5a^{2}-27ac$$

Short Solution Steps

Factor the expressions that are not already factored.

$$\frac{\frac{1}{6}b\left(5a+27c\right)a^{6}}{-\frac{1}{6}ba^{5}}$$

Cancel out ba^{5} in both numerator and denominator.

$$\frac{\frac{1}{6}a\left(5a+27c\right)}{-\frac{1}{6}}$$

Expand the expression.

$$\frac{\frac{5}{6}a^{2}+\frac{9}{2}ac}{-\frac{1}{6}}$$

Divide \frac{5}{6}a^{2}+\frac{9}{2}ac by -\frac{1}{6} by multiplying \frac{5}{6}a^{2}+\frac{9}{2}ac by the reciprocal of -\frac{1}{6}.

$$\frac{\left(\frac{5}{6}a^{2}+\frac{9}{2}ac\right)\times 6}{-1}$$

Anything divided by -1 gives its opposite.

$$-\left(\frac{5}{6}a^{2}+\frac{9}{2}ac\right)\times 6$$

Use the distributive property to multiply \frac{5}{6}a^{2}+\frac{9}{2}ac by 6.

$$-\left(5a^{2}+27ac\right)$$

To find the opposite of 5a^{2}+27ac, find the opposite of each term.

$$-5a^{2}-27ac$$