Solve -[(3/4)+(-2)]+(3/3)+(-1)^3+2*(2)-3)}

$$- [ ( \frac { 3 } { 4 } ) + ( - 2 ) ] + ( \frac { 3 } { 3 } ) + ( - 1 ) ^ { 3 } + 2 \cdot ( 2 ) - 3 ) \}$$

Evaluate

$$\frac{9}{4}=2.25$$

Solution Steps

Divide 3 by 3 to get 1.

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

Convert 2 to fraction \frac{8}{4}.

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

Since \frac{3}{4} and \frac{8}{4} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 8 from 3 to get -5.

$$-\left(-\frac{5}{4}\right)+1+\left(-1\right)^{3}+2\times 2-3$$

The opposite of -\frac{5}{4} is \frac{5}{4}.

$$\frac{5}{4}+1+\left(-1\right)^{3}+2\times 2-3$$

Convert 1 to fraction \frac{4}{4}.

$$\frac{5}{4}+\frac{4}{4}+\left(-1\right)^{3}+2\times 2-3$$

Since \frac{5}{4} and \frac{4}{4} have the same denominator, add them by adding their numerators.

$$\frac{5+4}{4}+\left(-1\right)^{3}+2\times 2-3$$

Add 5 and 4 to get 9.

$$\frac{9}{4}+\left(-1\right)^{3}+2\times 2-3$$

Calculate -1 to the power of 3 and get -1.

$$\frac{9}{4}-1+2\times 2-3$$

Convert 1 to fraction \frac{4}{4}.

$$\frac{9}{4}-\frac{4}{4}+2\times 2-3$$

Since \frac{9}{4} and \frac{4}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9-4}{4}+2\times 2-3$$

Subtract 4 from 9 to get 5.

$$\frac{5}{4}+2\times 2-3$$

Multiply 2 and 2 to get 4.

$$\frac{5}{4}+4-3$$

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

$$\frac{5}{4}+\frac{16}{4}-3$$

Since \frac{5}{4} and \frac{16}{4} have the same denominator, add them by adding their numerators.

$$\frac{5+16}{4}-3$$

Add 5 and 16 to get 21.

$$\frac{21}{4}-3$$

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

$$\frac{21}{4}-\frac{12}{4}$$

Since \frac{21}{4} and \frac{12}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{21-12}{4}$$

Subtract 12 from 21 to get 9.

$$\frac{9}{4}$$

Factor

$$\frac{3 ^ {2}}{2 ^ {2}} = 2\frac{1}{4} = 2.25$$