Solve (-3/4)*(-8+2/3-1/3)-3=(-2)^2

$$( - \frac { 3 } { 4 } ) \times ( - 8 + \frac { 2 } { 3 } - \frac { 1 } { 3 } ) - 3 = ( - 2 ) ^ { 2 }$$

Verify

Solution Steps

Convert -8 to fraction -\frac{24}{3}.

$$-\frac{3}{4}\left(-\frac{24}{3}+\frac{2}{3}-\frac{1}{3}\right)-3=\left(-2\right)^{2}$$

Since -\frac{24}{3} and \frac{2}{3} have the same denominator, add them by adding their numerators.

$$-\frac{3}{4}\left(\frac{-24+2}{3}-\frac{1}{3}\right)-3=\left(-2\right)^{2}$$

Add -24 and 2 to get -22.

$$-\frac{3}{4}\left(-\frac{22}{3}-\frac{1}{3}\right)-3=\left(-2\right)^{2}$$

Since -\frac{22}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{3}{4}\times \frac{-22-1}{3}-3=\left(-2\right)^{2}$$

Subtract 1 from -22 to get -23.

$$-\frac{3}{4}\left(-\frac{23}{3}\right)-3=\left(-2\right)^{2}$$

Multiply -\frac{3}{4} times -\frac{23}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-3\left(-23\right)}{4\times 3}-3=\left(-2\right)^{2}$$

Do the multiplications in the fraction \frac{-3\left(-23\right)}{4\times 3}.

$$\frac{69}{12}-3=\left(-2\right)^{2}$$

Reduce the fraction \frac{69}{12} to lowest terms by extracting and canceling out 3.

$$\frac{23}{4}-3=\left(-2\right)^{2}$$

Convert 3 to fraction \frac{12}{4}.

$$\frac{23}{4}-\frac{12}{4}=\left(-2\right)^{2}$$

Since \frac{23}{4} and \frac{12}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{23-12}{4}=\left(-2\right)^{2}$$

Subtract 12 from 23 to get 11.

$$\frac{11}{4}=\left(-2\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{11}{4}=4$$

Convert 4 to fraction \frac{16}{4}.

$$\frac{11}{4}=\frac{16}{4}$$

Compare \frac{11}{4} and \frac{16}{4}.

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