Solve =368quad[(3/28a^2b)(-7/4ab)-(-1/8a^3b)(+2b)]:(-1/4a^2b^2)

$$= 368 \quad [ ( \frac { 3 } { 28 } a ^ { 2 } b ) ( - \frac { 7 } { 4 } a b ) - ( - \frac { 1 } { 8 } a ^ { 3 } b ) ( + 2 b ) ] : ( - \frac { 1 } { 4 } a ^ { 2 } b ^ { 2 } )$$

Evaluate

$$-92a$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{368\left(\frac{3}{28}a^{3}b\left(-\frac{7}{4}\right)b-\left(-\frac{1}{8}a^{3}b\times 2b\right)\right)}{-\frac{1}{4}a^{2}b^{2}}$$

Multiply b and b to get b^{2}.

$$\frac{368\left(\frac{3}{28}a^{3}b^{2}\left(-\frac{7}{4}\right)-\left(-\frac{1}{8}a^{3}b\times 2b\right)\right)}{-\frac{1}{4}a^{2}b^{2}}$$

Multiply b and b to get b^{2}.

$$\frac{368\left(\frac{3}{28}a^{3}b^{2}\left(-\frac{7}{4}\right)-\left(-\frac{1}{8}a^{3}b^{2}\times 2\right)\right)}{-\frac{1}{4}a^{2}b^{2}}$$

Multiply \frac{3}{28} and -\frac{7}{4} to get -\frac{3}{16}.

$$\frac{368\left(-\frac{3}{16}a^{3}b^{2}-\left(-\frac{1}{8}a^{3}b^{2}\times 2\right)\right)}{-\frac{1}{4}a^{2}b^{2}}$$

Multiply -\frac{1}{8} and 2 to get -\frac{1}{4}.

$$\frac{368\left(-\frac{3}{16}a^{3}b^{2}-\left(-\frac{1}{4}a^{3}b^{2}\right)\right)}{-\frac{1}{4}a^{2}b^{2}}$$

The opposite of -\frac{1}{4}a^{3}b^{2} is \frac{1}{4}a^{3}b^{2}.

$$\frac{368\left(-\frac{3}{16}a^{3}b^{2}+\frac{1}{4}a^{3}b^{2}\right)}{-\frac{1}{4}a^{2}b^{2}}$$

Combine -\frac{3}{16}a^{3}b^{2} and \frac{1}{4}a^{3}b^{2} to get \frac{1}{16}a^{3}b^{2}.

$$\frac{368\times \frac{1}{16}a^{3}b^{2}}{-\frac{1}{4}a^{2}b^{2}}$$

Multiply 368 and \frac{1}{16} to get 23.

$$\frac{23a^{3}b^{2}}{-\frac{1}{4}a^{2}b^{2}}$$

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

$$\frac{23a}{-\frac{1}{4}}$$

Divide 23a by -\frac{1}{4} by multiplying 23a by the reciprocal of -\frac{1}{4}.

$$\frac{23a\times 4}{-1}$$

Multiply 23 and 4 to get 92.

$$\frac{92a}{-1}$$

Anything divided by -1 gives its opposite.

$$-92a$$

Expand

$$-92a$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\frac{368\left(\frac{3}{28}a^{3}b\left(-\frac{7}{4}\right)b-\left(-\frac{1}{8}a^{3}b\times 2b\right)\right)}{-\frac{1}{4}a^{2}b^{2}}$$

Multiply b and b to get b^{2}.

$$\frac{368\left(\frac{3}{28}a^{3}b^{2}\left(-\frac{7}{4}\right)-\left(-\frac{1}{8}a^{3}b\times 2b\right)\right)}{-\frac{1}{4}a^{2}b^{2}}$$

Multiply b and b to get b^{2}.

$$\frac{368\left(\frac{3}{28}a^{3}b^{2}\left(-\frac{7}{4}\right)-\left(-\frac{1}{8}a^{3}b^{2}\times 2\right)\right)}{-\frac{1}{4}a^{2}b^{2}}$$

Multiply \frac{3}{28} and -\frac{7}{4} to get -\frac{3}{16}.

$$\frac{368\left(-\frac{3}{16}a^{3}b^{2}-\left(-\frac{1}{8}a^{3}b^{2}\times 2\right)\right)}{-\frac{1}{4}a^{2}b^{2}}$$

Multiply -\frac{1}{8} and 2 to get -\frac{1}{4}.

$$\frac{368\left(-\frac{3}{16}a^{3}b^{2}-\left(-\frac{1}{4}a^{3}b^{2}\right)\right)}{-\frac{1}{4}a^{2}b^{2}}$$

The opposite of -\frac{1}{4}a^{3}b^{2} is \frac{1}{4}a^{3}b^{2}.

$$\frac{368\left(-\frac{3}{16}a^{3}b^{2}+\frac{1}{4}a^{3}b^{2}\right)}{-\frac{1}{4}a^{2}b^{2}}$$

Combine -\frac{3}{16}a^{3}b^{2} and \frac{1}{4}a^{3}b^{2} to get \frac{1}{16}a^{3}b^{2}.

$$\frac{368\times \frac{1}{16}a^{3}b^{2}}{-\frac{1}{4}a^{2}b^{2}}$$

Multiply 368 and \frac{1}{16} to get 23.

$$\frac{23a^{3}b^{2}}{-\frac{1}{4}a^{2}b^{2}}$$

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

$$\frac{23a}{-\frac{1}{4}}$$

Divide 23a by -\frac{1}{4} by multiplying 23a by the reciprocal of -\frac{1}{4}.

$$\frac{23a\times 4}{-1}$$

Multiply 23 and 4 to get 92.

$$\frac{92a}{-1}$$

Anything divided by -1 gives its opposite.

$$-92a$$