Solve 9+2*[-6-(-34):(6-14+14)+(-8)]-[32+2*(-18)+(-15)]

$$9 + 2 \cdot [ - 6 - ( - 34 ) : ( 6 - 14 + 14 ) + ( - 8 ) ] - [ 32 + 2 \cdot ( - 18 ) + ( - 15 ) ]$$

Evaluate

$$\frac{34}{3}\approx 11.333333333$$

Solution Steps

Subtract 14 from 6 to get -8.

$$9+2\left(-6-\frac{-34}{-8+14}-8\right)-\left(32+2\left(-18\right)-15\right)$$

Add -8 and 14 to get 6.

$$9+2\left(-6-\frac{-34}{6}-8\right)-\left(32+2\left(-18\right)-15\right)$$

Reduce the fraction \frac{-34}{6} to lowest terms by extracting and canceling out 2.

$$9+2\left(-6-\left(-\frac{17}{3}\right)-8\right)-\left(32+2\left(-18\right)-15\right)$$

The opposite of -\frac{17}{3} is \frac{17}{3}.

$$9+2\left(-6+\frac{17}{3}-8\right)-\left(32+2\left(-18\right)-15\right)$$

Convert -6 to fraction -\frac{18}{3}.

$$9+2\left(-\frac{18}{3}+\frac{17}{3}-8\right)-\left(32+2\left(-18\right)-15\right)$$

Since -\frac{18}{3} and \frac{17}{3} have the same denominator, add them by adding their numerators.

$$9+2\left(\frac{-18+17}{3}-8\right)-\left(32+2\left(-18\right)-15\right)$$

Add -18 and 17 to get -1.

$$9+2\left(-\frac{1}{3}-8\right)-\left(32+2\left(-18\right)-15\right)$$

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

$$9+2\left(-\frac{1}{3}-\frac{24}{3}\right)-\left(32+2\left(-18\right)-15\right)$$

Since -\frac{1}{3} and \frac{24}{3} have the same denominator, subtract them by subtracting their numerators.

$$9+2\times \frac{-1-24}{3}-\left(32+2\left(-18\right)-15\right)$$

Subtract 24 from -1 to get -25.

$$9+2\left(-\frac{25}{3}\right)-\left(32+2\left(-18\right)-15\right)$$

Express 2\left(-\frac{25}{3}\right) as a single fraction.

$$9+\frac{2\left(-25\right)}{3}-\left(32+2\left(-18\right)-15\right)$$

Multiply 2 and -25 to get -50.

$$9+\frac{-50}{3}-\left(32+2\left(-18\right)-15\right)$$

Fraction \frac{-50}{3} can be rewritten as -\frac{50}{3} by extracting the negative sign.

$$9-\frac{50}{3}-\left(32+2\left(-18\right)-15\right)$$

Convert 9 to fraction \frac{27}{3}.

$$\frac{27}{3}-\frac{50}{3}-\left(32+2\left(-18\right)-15\right)$$

Since \frac{27}{3} and \frac{50}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{27-50}{3}-\left(32+2\left(-18\right)-15\right)$$

Subtract 50 from 27 to get -23.

$$-\frac{23}{3}-\left(32+2\left(-18\right)-15\right)$$

Multiply 2 and -18 to get -36.

$$-\frac{23}{3}-\left(32-36-15\right)$$

Subtract 36 from 32 to get -4.

$$-\frac{23}{3}-\left(-4-15\right)$$

Subtract 15 from -4 to get -19.

$$-\frac{23}{3}-\left(-19\right)$$

The opposite of -19 is 19.

$$-\frac{23}{3}+19$$

Convert 19 to fraction \frac{57}{3}.

$$-\frac{23}{3}+\frac{57}{3}$$

Since -\frac{23}{3} and \frac{57}{3} have the same denominator, add them by adding their numerators.

$$\frac{-23+57}{3}$$

Add -23 and 57 to get 34.

$$\frac{34}{3}$$

Factor

$$\frac{2 \cdot 17}{3} = 11\frac{1}{3} = 11.333333333333334$$