Solve 3/5*2/9+55/9=35/18-(frac{3}{1}+frac{12}{3}*frac{12}{10})*frac{5}{23}

$$\frac { 3 } { 5 } \times \frac { 2 } { 9 } + \frac { 55 } { 9 } = \frac { 35 } { 18 } - ( \frac { 3 } { 1 } + \frac { 12 } { 3 } \times \frac { 12 } { 10 } ) \times \frac { 5 } { 23 }$$

Verify

Solution Steps

Multiply \frac{3}{5} times \frac{2}{9} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3\times 2}{5\times 9}+\frac{55}{9}=\frac{35}{18}-\left(\frac{3}{1}+\frac{12}{3}\times \frac{12}{10}\right)\times \frac{5}{23}$$

Do the multiplications in the fraction \frac{3\times 2}{5\times 9}.

$$\frac{6}{45}+\frac{55}{9}=\frac{35}{18}-\left(\frac{3}{1}+\frac{12}{3}\times \frac{12}{10}\right)\times \frac{5}{23}$$

Reduce the fraction \frac{6}{45} to lowest terms by extracting and canceling out 3.

$$\frac{2}{15}+\frac{55}{9}=\frac{35}{18}-\left(\frac{3}{1}+\frac{12}{3}\times \frac{12}{10}\right)\times \frac{5}{23}$$

Least common multiple of 15 and 9 is 45. Convert \frac{2}{15} and \frac{55}{9} to fractions with denominator 45.

$$\frac{6}{45}+\frac{275}{45}=\frac{35}{18}-\left(\frac{3}{1}+\frac{12}{3}\times \frac{12}{10}\right)\times \frac{5}{23}$$

Since \frac{6}{45} and \frac{275}{45} have the same denominator, add them by adding their numerators.

$$\frac{6+275}{45}=\frac{35}{18}-\left(\frac{3}{1}+\frac{12}{3}\times \frac{12}{10}\right)\times \frac{5}{23}$$

Add 6 and 275 to get 281.

$$\frac{281}{45}=\frac{35}{18}-\left(\frac{3}{1}+\frac{12}{3}\times \frac{12}{10}\right)\times \frac{5}{23}$$

Anything divided by one gives itself.

$$\frac{281}{45}=\frac{35}{18}-\left(3+\frac{12}{3}\times \frac{12}{10}\right)\times \frac{5}{23}$$

Divide 12 by 3 to get 4.

$$\frac{281}{45}=\frac{35}{18}-\left(3+4\times \frac{12}{10}\right)\times \frac{5}{23}$$

Reduce the fraction \frac{12}{10} to lowest terms by extracting and canceling out 2.

$$\frac{281}{45}=\frac{35}{18}-\left(3+4\times \frac{6}{5}\right)\times \frac{5}{23}$$

Express 4\times \frac{6}{5} as a single fraction.

$$\frac{281}{45}=\frac{35}{18}-\left(3+\frac{4\times 6}{5}\right)\times \frac{5}{23}$$

Multiply 4 and 6 to get 24.

$$\frac{281}{45}=\frac{35}{18}-\left(3+\frac{24}{5}\right)\times \frac{5}{23}$$

Convert 3 to fraction \frac{15}{5}.

$$\frac{281}{45}=\frac{35}{18}-\left(\frac{15}{5}+\frac{24}{5}\right)\times \frac{5}{23}$$

Since \frac{15}{5} and \frac{24}{5} have the same denominator, add them by adding their numerators.

$$\frac{281}{45}=\frac{35}{18}-\frac{15+24}{5}\times \frac{5}{23}$$

Add 15 and 24 to get 39.

$$\frac{281}{45}=\frac{35}{18}-\frac{39}{5}\times \frac{5}{23}$$

Multiply \frac{39}{5} times \frac{5}{23} by multiplying numerator times numerator and denominator times denominator.

$$\frac{281}{45}=\frac{35}{18}-\frac{39\times 5}{5\times 23}$$

Cancel out 5 in both numerator and denominator.

$$\frac{281}{45}=\frac{35}{18}-\frac{39}{23}$$

Least common multiple of 18 and 23 is 414. Convert \frac{35}{18} and \frac{39}{23} to fractions with denominator 414.

$$\frac{281}{45}=\frac{805}{414}-\frac{702}{414}$$

Since \frac{805}{414} and \frac{702}{414} have the same denominator, subtract them by subtracting their numerators.

$$\frac{281}{45}=\frac{805-702}{414}$$

Subtract 702 from 805 to get 103.

$$\frac{281}{45}=\frac{103}{414}$$

Least common multiple of 45 and 414 is 2070. Convert \frac{281}{45} and \frac{103}{414} to fractions with denominator 2070.

$$\frac{12926}{2070}=\frac{515}{2070}$$

Compare \frac{12926}{2070} and \frac{515}{2070}.

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