Solve (1/6*9)-(3)^2+9/2

$$( \frac { 1 } { 6 } \cdot 9 ) - ( 3 ) ^ { 2 } + \frac { 9 } { 2 }$$

Evaluate

$$-3$$

Solution Steps

Multiply \frac{1}{6} and 9 to get \frac{9}{6}.

$$\frac{9}{6}-3^{2}+\frac{9}{2}$$

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

$$\frac{3}{2}-3^{2}+\frac{9}{2}$$

Calculate 3 to the power of 2 and get 9.

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

Convert 9 to fraction \frac{18}{2}.

$$\frac{3}{2}-\frac{18}{2}+\frac{9}{2}$$

Since \frac{3}{2} and \frac{18}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3-18}{2}+\frac{9}{2}$$

Subtract 18 from 3 to get -15.

$$-\frac{15}{2}+\frac{9}{2}$$

Since -\frac{15}{2} and \frac{9}{2} have the same denominator, add them by adding their numerators.

$$\frac{-15+9}{2}$$

Add -15 and 9 to get -6.

$$\frac{-6}{2}$$

Divide -6 by 2 to get -3.

$$-3$$

Factor

$$-3$$