Solve 1/6+3+6/(1+3)(-6-3)+5/6-3+1/1

$$\frac { 1 } { 6 + 3 } + \frac { 6 } { ( 1 + 3 ) ( - 6 - 3 ) } + \frac { 5 } { 6 - 3 } + \frac { 1 } { 1 }$$

Evaluate

$$\frac{47}{18}\approx 2.611111111$$

Solution Steps

Divide 1 by 1 to get 1.

$$\frac{1}{6+3}+\frac{6}{\left(1+3\right)\left(-6-3\right)}+\frac{5}{6-3}+1$$

Add 6 and 3 to get 9.

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

Add 1 and 3 to get 4.

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

Subtract 3 from -6 to get -9.

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

Multiply 4 and -9 to get -36.

$$\frac{1}{9}+\frac{6}{-36}+\frac{5}{6-3}+1$$

Reduce the fraction \frac{6}{-36} to lowest terms by extracting and canceling out 6.

$$\frac{1}{9}-\frac{1}{6}+\frac{5}{6-3}+1$$

Least common multiple of 9 and 6 is 18. Convert \frac{1}{9} and \frac{1}{6} to fractions with denominator 18.

$$\frac{2}{18}-\frac{3}{18}+\frac{5}{6-3}+1$$

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

$$\frac{2-3}{18}+\frac{5}{6-3}+1$$

Subtract 3 from 2 to get -1.

$$-\frac{1}{18}+\frac{5}{6-3}+1$$

Subtract 3 from 6 to get 3.

$$-\frac{1}{18}+\frac{5}{3}+1$$

Least common multiple of 18 and 3 is 18. Convert -\frac{1}{18} and \frac{5}{3} to fractions with denominator 18.

$$-\frac{1}{18}+\frac{30}{18}+1$$

Since -\frac{1}{18} and \frac{30}{18} have the same denominator, add them by adding their numerators.

$$\frac{-1+30}{18}+1$$

Add -1 and 30 to get 29.

$$\frac{29}{18}+1$$

Convert 1 to fraction \frac{18}{18}.

$$\frac{29}{18}+\frac{18}{18}$$

Since \frac{29}{18} and \frac{18}{18} have the same denominator, add them by adding their numerators.

$$\frac{29+18}{18}$$

Add 29 and 18 to get 47.

$$\frac{47}{18}$$

Factor

$$\frac{47}{2 \cdot 3 ^ {2}} = 2\frac{11}{18} = 2.611111111111111$$