Solve 17/91+(2/3-1/5)-1/3*(-4)^2

$$1 \frac { 7 } { 9 } 1 + ( \frac { 2 } { 3 } - \frac { 1 } { 5 } ) - \frac { 1 } { 3 } \times ( - 4 ) ^ { 2 }$$

Evaluate

$$-\frac{139}{45}\approx -3.088888889$$

Solution Steps

Multiply 1 and 9 to get 9.

$$\frac{9+7}{9}\times 1+\frac{2}{3}-\frac{1}{5}-\frac{1}{3}\left(-4\right)^{2}$$

Add 9 and 7 to get 16.

$$\frac{16}{9}\times 1+\frac{2}{3}-\frac{1}{5}-\frac{1}{3}\left(-4\right)^{2}$$

Multiply \frac{16}{9} and 1 to get \frac{16}{9}.

$$\frac{16}{9}+\frac{2}{3}-\frac{1}{5}-\frac{1}{3}\left(-4\right)^{2}$$

Least common multiple of 9 and 3 is 9. Convert \frac{16}{9} and \frac{2}{3} to fractions with denominator 9.

$$\frac{16}{9}+\frac{6}{9}-\frac{1}{5}-\frac{1}{3}\left(-4\right)^{2}$$

Since \frac{16}{9} and \frac{6}{9} have the same denominator, add them by adding their numerators.

$$\frac{16+6}{9}-\frac{1}{5}-\frac{1}{3}\left(-4\right)^{2}$$

Add 16 and 6 to get 22.

$$\frac{22}{9}-\frac{1}{5}-\frac{1}{3}\left(-4\right)^{2}$$

Least common multiple of 9 and 5 is 45. Convert \frac{22}{9} and \frac{1}{5} to fractions with denominator 45.

$$\frac{110}{45}-\frac{9}{45}-\frac{1}{3}\left(-4\right)^{2}$$

Since \frac{110}{45} and \frac{9}{45} have the same denominator, subtract them by subtracting their numerators.

$$\frac{110-9}{45}-\frac{1}{3}\left(-4\right)^{2}$$

Subtract 9 from 110 to get 101.

$$\frac{101}{45}-\frac{1}{3}\left(-4\right)^{2}$$

Calculate -4 to the power of 2 and get 16.

$$\frac{101}{45}-\frac{1}{3}\times 16$$

Multiply \frac{1}{3} and 16 to get \frac{16}{3}.

$$\frac{101}{45}-\frac{16}{3}$$

Least common multiple of 45 and 3 is 45. Convert \frac{101}{45} and \frac{16}{3} to fractions with denominator 45.

$$\frac{101}{45}-\frac{240}{45}$$

Since \frac{101}{45} and \frac{240}{45} have the same denominator, subtract them by subtracting their numerators.

$$\frac{101-240}{45}$$

Subtract 240 from 101 to get -139.

$$-\frac{139}{45}$$

Factor

$$-\frac{139}{45} = -3\frac{4}{45} = -3.088888888888889$$